{"id":15381,"date":"2021-10-11T23:05:41","date_gmt":"2021-10-11T23:05:41","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/read-or-watch-the-quadratic-formula\/"},"modified":"2021-10-18T03:32:47","modified_gmt":"2021-10-18T03:32:47","slug":"quadratic-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/quadratic-formula\/","title":{"raw":"Quadratic Formula","rendered":"Quadratic Formula"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write a quadratic equation in standard form and identify the values of\u00a0[latex]a[\/latex],\u00a0[latex]b[\/latex], and [latex]c[\/latex] in a\u00a0standard form quadratic equation.<\/li>\r\n \t<li>Use the Quadratic Formula to find all real solutions of a quadratic equation<\/li>\r\n<\/ul>\r\n<\/div>\r\nYou can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for [latex]x[\/latex], you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.\r\n\r\n[caption id=\"attachment_5188\" align=\"aligncenter\" width=\"465\"]<img class=\" wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/24025157\/Screen-Shot-2016-06-23-at-7.51.28-PM-300x216.png\" alt=\"x is a fraction with the numerator -b plus or minus the square root of b squared minus 4 times a times c and the denominator is 2 times a\" width=\"465\" height=\"335\" \/> Quadratic formula[\/caption]\r\n\r\n&nbsp;\r\n\r\nWe can derive the quadratic formula by completing the square. First,\u00a0assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:\r\n<ol>\r\n \t<li>First, move the constant term to the right side of the equal sign:\r\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div><\/li>\r\n \t<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\cdot\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\r\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div><\/li>\r\n \t<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\r\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div><\/li>\r\n \t<li>Now, use the square root property, which gives\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div><\/li>\r\n \t<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\r\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div><\/li>\r\n<\/ol>\r\nThis formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].\r\n<h2>Writing a Quadratic Equation in Standard Form<\/h2>\r\nRemember that the form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it is\u00a0<i>vital<\/i> that you be sure the equation is in this form. If you do not, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.\r\n\r\nThe following examples show how to ensure that your quadratic equation is in standard form and then correctly identify the values you will be using for a, b, and c in the Quadratic Formula.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRewrite the equation [latex]3x+2x^{2}+4=5[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.\r\n\r\n[reveal-answer q=\"489648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"489648\"]First be sure that the right side of the equation is 0. In this case, all you need to do is subtract [latex]5[\/latex] from both sides.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}3x+2x^{2}+4=5\\\\3x+2x^{2}+4\u20135=5\u20135\\end{array}[\/latex]<\/p>\r\nSimplify, and write the terms with the exponent on the variable in descending order.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+2x^{2}-1=0\\\\2x^{2}+3x-1=0\\end{array}[\/latex]<\/p>\r\nNow that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant. Note that since the constant 1 is subtracted, <i>c <\/i>must be negative.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,3x\\,\\,\\,-\\,\\,\\,1\\,\\,\\,=\\,\\,\\,0\\\\\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,ax^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,bx\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,a=2,\\,\\,b=3,\\,\\,c=-1\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]2x^{2}+3x\u20131=0;a=2,b=3,c=\u22121[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nRewrite the equation [latex]2(x+3)^{2}\u20135x=6[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.\r\n\r\n[reveal-answer q=\"585220\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"585220\"]First be sure that the right side of the equation is [latex]0[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2\\left(x+3\\right)^{2}\u20135x=6\\\\2(x+3)^{2}\u20135x\u20136=6\u20136\\end{array}[\/latex]<\/p>\r\nExpand the squared binomial, then simplify by combining like terms.\r\n\r\nBe sure to write the terms with the exponent on the variable in descending order.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(x^{2}+6x+9\\right)-5x-6=0\\\\2x^{2}+12x+18\u20135x\u20136=0\\\\2x^{2}+12x\u20135x+18\u20136=0\\\\2x^{2}+7x+12=0\\end{array}[\/latex]<\/p>\r\nNow that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,7x\\,\\,\\,+\\,\\,\\,12\\,\\,\\,=\\,\\,\\,0\\\\\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,\\,a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,\\,\\,\\,\\,a=2,\\,\\,b=7,\\,\\,c=7\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]2x^{2}+7x+12=0;\\,\\,a=2,b=7,c=12[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]91600[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\r\nThe Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.\r\n<ol>\r\n \t<li>Put the equation in standard form first.<\/li>\r\n \t<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be sure to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\r\n \t<li>Carefully substitute the values noted in step\u00a0[latex]2[\/latex] into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\r\n \t<li>Simplify as much as possible.<\/li>\r\n \t<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added and one in which it is subtracted<i>.<\/i><\/li>\r\n \t<li>Simplify both values to get the possible solutions.<\/li>\r\n<\/ol>\r\nThat is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will go back and solve it again using another factoring method.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].\r\n\r\n[reveal-answer q=\"296770\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296770\"]\r\n\r\nFirst write the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1, b=4, c=-5\\end{array}[\/latex]<\/p>\r\nNote that the subtraction sign means the constant <i>c<\/i> is negative.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\u00a0[latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify, being careful to get the signs correct.\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\r\nSimplify some more.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\r\nSimplify the radical: [latex] \\sqrt{36}=6[\/latex].\r\n<p style=\"text-align: center;\">[latex] x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\r\nSeparate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other,\u00a0[latex]6[\/latex] is subtracted.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\r\nThe solutions are [latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nYou can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\r\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nYou get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You have solved the equation successfully using the Quadratic Formula!\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n[ohm_question]4015[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nSometimes, it may be easier to solve an equation using conventional factoring methods like finding number pairs that sum to one number (in this example, [latex]4[\/latex]) and that produce a specific product (in this example [latex]\u22125[\/latex]) when multiplied. The power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.\r\n\r\nIn our next two video examples, we will see, first, a quadratic equation with two real, rational solutions and, second, a quadratic equation that has irrational solutions and that could not have been solved using factoring.\r\n\r\nhttps:\/\/youtu.be\/xtwO-n8lRPw\r\n\r\nIn the next video example, we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.\r\n\r\nhttps:\/\/youtu.be\/tF0muV86dr0\r\n\r\nMost of the quadratic equations you have looked at have two solutions, like the one above. The following example is a little different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].\r\n\r\n[reveal-answer q=\"998241\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"998241\"]\r\n\r\nSubtract [latex]6[\/latex]<i>x <\/i>from each side and add\u00a0[latex]16[\/latex] to both sides to put the equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\r\nIdentify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\r\nSubstitute the values into the Quadratic Formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\r\nSince the square root of\u00a0[latex]0[\/latex] is\u00a0[latex]0[\/latex], and both adding and subtracting\u00a0[latex]0[\/latex] give the same result, there is only one possible value.\r\n<p style=\"text-align: center;\">[latex] x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\r\nThe answer is [latex]x=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAgain, check using the original equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nIn the following video we show an example of using the quadratic formula to solve a quadratic\u00a0equation that has one repeated solution.\r\n\r\nhttps:\/\/youtu.be\/OXwwzWcxFgE\r\n\r\nIn the next example, we will show that some quadratic equations do not have real solutions. \u00a0As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[\/latex]\r\n[reveal-answer q=\"268005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"268005\"]\r\n\r\nAdd [latex]x[\/latex] to both sides and add 3 to both sides to get the quadratic equation in standard form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+x=-x-3\\\\x^{2}+2x+3=0\\end{array}[\/latex]<\/p>\r\nIdentify a, b, c.\r\n<p style=\"text-align: center;\">[latex]a=1, b=2, c=3[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Substitute values for a, b, c into the quadratic formula.<\/p>\r\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Simplify<\/p>\r\n<p style=\"text-align: left;\">[latex] x=\\frac{-2\\pm \\sqrt{-8}}{2}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>Summary<\/h3>\r\nThe Quadratic Formula is a useful way to solve any quadratic equation. The Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex] [\/latex].\u00a0When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.\r\n<h2>Contribute!<\/h2>\r\n<div style=\"margin-bottom: 8px;\">Did you have an idea for improving this content? We\u2019d love your input.<\/div>\r\n<a style=\"font-size: 10pt; font-weight: 600; color: #077fab; text-decoration: none; border: 2px solid #077fab; border-radius: 7px; padding: 5px 25px; text-align: center; cursor: pointer; line-height: 1.5em;\" href=\"https:\/\/docs.google.com\/document\/d\/1MGLJUw45yPf5ymtqNavz25oZ4cXnk4rA2w3CnfgVycw\" target=\"_blank\" rel=\"noopener\">Improve this page<\/a><a style=\"margin-left: 16px;\" href=\"https:\/\/docs.google.com\/document\/d\/1vy-T6DtTF-BbMfpVEI7VP_R7w2A4anzYZLXR8Pk4Fu4\" target=\"_blank\" rel=\"noopener\">Learn More<\/a>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write a quadratic equation in standard form and identify the values of\u00a0[latex]a[\/latex],\u00a0[latex]b[\/latex], and [latex]c[\/latex] in a\u00a0standard form quadratic equation.<\/li>\n<li>Use the Quadratic Formula to find all real solutions of a quadratic equation<\/li>\n<\/ul>\n<\/div>\n<p>You can solve any quadratic equation by <strong>completing the square<\/strong>\u2014rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[\/latex]\u00a0and then solve for [latex]x[\/latex], you find that [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. This equation is known as the Quadratic Formula.<\/p>\n<div id=\"attachment_5188\" style=\"width: 475px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5188\" class=\"wp-image-5188\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/24025157\/Screen-Shot-2016-06-23-at-7.51.28-PM-300x216.png\" alt=\"x is a fraction with the numerator -b plus or minus the square root of b squared minus 4 times a times c and the denominator is 2 times a\" width=\"465\" height=\"335\" \/><\/p>\n<p id=\"caption-attachment-5188\" class=\"wp-caption-text\">Quadratic formula<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We can derive the quadratic formula by completing the square. First,\u00a0assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[\/latex] and obtain a positive <em>a<\/em>. Given [latex]a{x}^{2}+bx+c=0[\/latex], [latex]a\\ne 0[\/latex], we will complete the square as follows:<\/p>\n<ol>\n<li>First, move the constant term to the right side of the equal sign:\n<div style=\"text-align: center;\">[latex]a{x}^{2}+bx=-c[\/latex]<\/div>\n<\/li>\n<li>As we want the leading coefficient to equal 1, divide through by <em>a<\/em>:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x=-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Then, find [latex]\\frac{1}{2}[\/latex] of the middle term, and add [latex]{\\left(\\frac{1}{2}\\cdot\\frac{b}{a}\\right)}^{2}=\\frac{{b}^{2}}{4{a}^{2}}[\/latex] to both sides of the equal sign:\n<div style=\"text-align: center;\">[latex]{x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}}=\\frac{{b}^{2}}{4{a}^{2}}-\\frac{c}{a}[\/latex]<\/div>\n<\/li>\n<li>Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:\n<div style=\"text-align: center;\">[latex]{\\left(x+\\frac{b}{2a}\\right)}^{2}=\\frac{{b}^{2}-4ac}{4{a}^{2}}[\/latex]<\/div>\n<\/li>\n<li>Now, use the square root property, which gives\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}x+\\frac{b}{2a}=\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x+\\frac{b}{2a}=\\frac{\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Finally, add [latex]-\\frac{b}{2a}[\/latex] to both sides of the equation and combine the terms on the right side. Thus,\n<div style=\"text-align: center;\">[latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p>This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[\/latex].<\/p>\n<h2>Writing a Quadratic Equation in Standard Form<\/h2>\n<p>Remember that the form [latex]ax^{2}+bx+c=0[\/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it is\u00a0<i>vital<\/i> that you be sure the equation is in this form. If you do not, you might use the wrong values for <i>a<\/i>, <i>b<\/i>, or <i>c<\/i>, and then the formula will give incorrect solutions.<\/p>\n<p>The following examples show how to ensure that your quadratic equation is in standard form and then correctly identify the values you will be using for a, b, and c in the Quadratic Formula.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rewrite the equation [latex]3x+2x^{2}+4=5[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q489648\">Show Solution<\/span><\/p>\n<div id=\"q489648\" class=\"hidden-answer\" style=\"display: none\">First be sure that the right side of the equation is 0. In this case, all you need to do is subtract [latex]5[\/latex] from both sides.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}3x+2x^{2}+4=5\\\\3x+2x^{2}+4\u20135=5\u20135\\end{array}[\/latex]<\/p>\n<p>Simplify, and write the terms with the exponent on the variable in descending order.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}3x+2x^{2}-1=0\\\\2x^{2}+3x-1=0\\end{array}[\/latex]<\/p>\n<p>Now that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant. Note that since the constant 1 is subtracted, <i>c <\/i>must be negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,3x\\,\\,\\,-\\,\\,\\,1\\,\\,\\,=\\,\\,\\,0\\\\\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,ax^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,bx\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,a=2,\\,\\,b=3,\\,\\,c=-1\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2x^{2}+3x\u20131=0;a=2,b=3,c=\u22121[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Rewrite the equation [latex]2(x+3)^{2}\u20135x=6[\/latex]\u00a0in standard form and identify <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q585220\">Show Solution<\/span><\/p>\n<div id=\"q585220\" class=\"hidden-answer\" style=\"display: none\">First be sure that the right side of the equation is [latex]0[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}2\\left(x+3\\right)^{2}\u20135x=6\\\\2(x+3)^{2}\u20135x\u20136=6\u20136\\end{array}[\/latex]<\/p>\n<p>Expand the squared binomial, then simplify by combining like terms.<\/p>\n<p>Be sure to write the terms with the exponent on the variable in descending order.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2\\left(x^{2}+6x+9\\right)-5x-6=0\\\\2x^{2}+12x+18\u20135x\u20136=0\\\\2x^{2}+12x\u20135x+18\u20136=0\\\\2x^{2}+7x+12=0\\end{array}[\/latex]<\/p>\n<p>Now that the equation is in standard form, you can read the values of <i>a<\/i>, <i>b<\/i>, and <i>c<\/i> from the coefficients and constant.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^{2}\\,\\,\\,+\\,\\,\\,7x\\,\\,\\,+\\,\\,\\,12\\,\\,\\,=\\,\\,\\,0\\\\\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\\\\\,\\,a\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,c\\\\\\\\\\,\\,\\,\\,\\,\\,a=2,\\,\\,b=7,\\,\\,c=7\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]2x^{2}+7x+12=0;\\,\\,a=2,b=7,c=12[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm91600\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=91600&theme=oea&iframe_resize_id=ohm91600&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Solving a Quadratic Equation using the Quadratic Formula<\/h2>\n<p>The Quadratic Formula will work with <i>any<\/i> quadratic equation, but <i>only<\/i> if the equation is in standard form, [latex]ax^{2}+bx+c=0[\/latex]. To use it, follow these steps.<\/p>\n<ol>\n<li>Put the equation in standard form first.<\/li>\n<li>Identify the coefficients, <i>a<\/i>, <i>b,<\/i> and <i>c. <\/i>Be sure to include negative signs if the <i>bx<\/i> or <i>c<\/i> terms are subtracted.<\/li>\n<li>Carefully substitute the values noted in step\u00a0[latex]2[\/latex] into the equation. To avoid needless errors, use parentheses around each number input into the formula.<\/li>\n<li>Simplify as much as possible.<\/li>\n<li>Use the [latex]\\pm[\/latex] in front of the radical to separate the solution into two values: one in which the square root is added and one in which it is subtracted<i>.<\/i><\/li>\n<li>Simplify both values to get the possible solutions.<\/li>\n<\/ol>\n<p>That is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will go back and solve it again using another factoring method.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q296770\">Show Solution<\/span><\/p>\n<div id=\"q296770\" class=\"hidden-answer\" style=\"display: none\">\n<p>First write the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}+4x=5\\,\\,\\,\\\\x^{2}+4x-5=0\\,\\,\\,\\\\\\\\a=1, b=4, c=-5\\end{array}[\/latex]<\/p>\n<p>Note that the subtraction sign means the constant <i>c<\/i> is negative.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,+\\,\\,\\,4x\\,\\,\\,-\\,\\,\\,5\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,c\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.\u00a0[latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\x=\\frac{-4\\pm \\sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify, being careful to get the signs correct.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm\\sqrt{16+20}}{2}[\/latex]<\/p>\n<p>Simplify some more.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm \\sqrt{36}}{2}[\/latex]<\/p>\n<p>Simplify the radical: [latex]\\sqrt{36}=6[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-4\\pm 6}{2}[\/latex]<\/p>\n<p>Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other,\u00a0[latex]6[\/latex] is subtracted.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x=\\frac{-4+6}{2}=\\frac{2}{2}=1\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-4-6}{2}=\\frac{-10}{2}=-5\\end{array}[\/latex]<\/p>\n<p>The solutions are [latex]x=1\\,\\,\\,\\text{or}\\,\\,\\,-5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>You can check these solutions by substituting [latex]1[\/latex] and [latex]\u22125[\/latex] into the original equation.<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=1\\\\x^{2}+4x=5\\\\\\left(1\\right)^{2}+4\\left(1\\right)=5\\\\1+4=5\\\\5=5\\end{array}[\/latex]<\/td>\n<td style=\"text-align: center;\">[latex]\\begin{array}{r}x=-5\\\\x^{2}+4x=5\\,\\,\\,\\,\\,\\\\\\left(-5\\right)^{2}+4\\left(-5\\right)=5\\,\\,\\,\\,\\,\\\\25-20=5\\,\\,\\,\\,\\,\\\\5=5\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>You get two true statements, so you know that both solutions work: [latex]x=1[\/latex] or [latex]-5[\/latex]. You have solved the equation successfully using the Quadratic Formula!<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4015\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4015&theme=oea&iframe_resize_id=ohm4015&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Sometimes, it may be easier to solve an equation using conventional factoring methods like finding number pairs that sum to one number (in this example, [latex]4[\/latex]) and that produce a specific product (in this example [latex]\u22125[\/latex]) when multiplied. The power of the Quadratic Formula is that it can be used to solve <i>any<\/i> quadratic equation, even those where finding number combinations will not work.<\/p>\n<p>In our next two video examples, we will see, first, a quadratic equation with two real, rational solutions and, second, a quadratic equation that has irrational solutions and that could not have been solved using factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Quadratic Formula - Two Real Rational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xtwO-n8lRPw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next video example, we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex2:  Quadratic Formula - Two Real Irrational Solutions\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tF0muV86dr0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Most of the quadratic equations you have looked at have two solutions, like the one above. The following example is a little different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q998241\">Show Solution<\/span><\/p>\n<div id=\"q998241\" class=\"hidden-answer\" style=\"display: none\">\n<p>Subtract [latex]6[\/latex]<i>x <\/i>from each side and add\u00a0[latex]16[\/latex] to both sides to put the equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}-2x=6x-16\\\\x^{2}-2x-6x+16=0\\\\x^{2}-8x+16=0\\end{array}[\/latex]<\/p>\n<p>Identify the coefficients <i>a<\/i>, <i>b<\/i>, and <i>c<\/i>. [latex]x^{2}=1x^{2}[\/latex], so [latex]a=1[\/latex].<i> <\/i>Since [latex]8x[\/latex]\u00a0is subtracted, <i>b<\/i> is negative.\u00a0[latex]a=1,b=-8,c=16[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{x}^{2}}\\,\\,\\,-\\,\\,\\,8x\\,\\,\\,+\\,\\,\\,16\\,\\,\\,=\\,\\,\\,0\\\\\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\downarrow\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\a{{x}^{2}}\\,\\,\\,+\\,\\,\\,bx\\,\\,\\,+\\,\\,\\,\\,c\\,\\,\\,\\,=\\,\\,\\,0\\end{array}[\/latex]<\/p>\n<p>Substitute the values into the Quadratic Formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-(-8)\\pm \\sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{64-64}}{2}[\/latex]<\/p>\n<p>Since the square root of\u00a0[latex]0[\/latex] is\u00a0[latex]0[\/latex], and both adding and subtracting\u00a0[latex]0[\/latex] give the same result, there is only one possible value.<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{8\\pm \\sqrt{0}}{2}=\\frac{8}{2}=4[\/latex]<\/p>\n<p>The answer is [latex]x=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Again, check using the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x^{2}-2x=6x-16\\,\\,\\,\\,\\,\\\\\\left(4\\right)^{2}-2\\left(4\\right)=6\\left(4\\right)-16\\\\16-8=24-16\\,\\,\\,\\,\\,\\,\\\\8=8\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>In the following video we show an example of using the quadratic formula to solve a quadratic\u00a0equation that has one repeated solution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Quadratic Formula - One Real Rational Repeated Solution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/OXwwzWcxFgE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will show that some quadratic equations do not have real solutions. \u00a0As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the Quadratic Formula to solve the equation [latex]x^2+x=-x-3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q268005\">Show Solution<\/span><\/p>\n<div id=\"q268005\" class=\"hidden-answer\" style=\"display: none\">\n<p>Add [latex]x[\/latex] to both sides and add 3 to both sides to get the quadratic equation in standard form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+x=-x-3\\\\x^{2}+2x+3=0\\end{array}[\/latex]<\/p>\n<p>Identify a, b, c.<\/p>\n<p style=\"text-align: center;\">[latex]a=1, b=2, c=3[\/latex]<\/p>\n<p style=\"text-align: left;\">Substitute values for a, b, c into the quadratic formula.<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-2\\pm \\sqrt{{{(2)}^{2}}-4(1)(3)}}{2(1)}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify<\/p>\n<p style=\"text-align: left;\">[latex]x=\\frac{-2\\pm \\sqrt{-8}}{2}[\/latex]<\/p>\n<p style=\"text-align: left;\">Since the square root of a negative number is not defined for real numbers, there are no real number solutions to this equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Summary<\/h3>\n<p>The Quadratic Formula is a useful way to solve any quadratic equation. The Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex], is found by completing the square of the quadratic equation [latex][\/latex].\u00a0When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation.<\/p>\n<h2>Contribute!<\/h2>\n<div style=\"margin-bottom: 8px;\">Did you have an idea for improving this content? We\u2019d love your input.<\/div>\n<p><a style=\"font-size: 10pt; font-weight: 600; color: #077fab; text-decoration: none; border: 2px solid #077fab; border-radius: 7px; padding: 5px 25px; text-align: center; cursor: pointer; line-height: 1.5em;\" href=\"https:\/\/docs.google.com\/document\/d\/1MGLJUw45yPf5ymtqNavz25oZ4cXnk4rA2w3CnfgVycw\" target=\"_blank\" rel=\"noopener\">Improve this page<\/a><a style=\"margin-left: 16px;\" href=\"https:\/\/docs.google.com\/document\/d\/1vy-T6DtTF-BbMfpVEI7VP_R7w2A4anzYZLXR8Pk4Fu4\" target=\"_blank\" rel=\"noopener\">Learn More<\/a><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-15381\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Quadratic Formula Application - Time for an Object to Hit the Ground. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RcVeuJhcuL0\">https:\/\/youtu.be\/RcVeuJhcuL0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Quadratic Formula Application - Determine the Width of a Border. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Zxe-SdwutxA\">https:\/\/youtu.be\/Zxe-SdwutxA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex2: Quadratic Formula - Two Real Irrational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tF0muV86dr0\">https:\/\/youtu.be\/tF0muV86dr0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Quadratic Formula - Two Real Rational Solutions. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xtwO-n8lRPw\">https:\/\/youtu.be\/xtwO-n8lRPw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":167848,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex2: Quadratic Formula - Two Real Irrational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tF0muV86dr0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Quadratic Formula - Two Real Rational Solutions\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xtwO-n8lRPw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Quadratic Formula Application - 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