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A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2.^{[1] X Research source } There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.
Steps
Method 1
Method 1 of 3:
Factoring the Equation

1Combine all of the like terms and move them to one side of the equation. The first step to factoring an equation is to move all of the terms to one side of the equation, keeping the term positive. To combine the terms, add or subtract all of the terms, the terms, and the constants (integer terms), moving them to one side of the equation so that nothing remains on the other side. Once the other side has no remaining terms, you can just write "0" on that side of the equal sign. Here's how you do it:^{[2] X Research source }

2Factor the expression. To factor the expression, you have to use the factors of the term (3), and the factors of the constant term (4), to make them multiply and then add up to the middle term, (11). Here's how you do it:
 Since only has one set of possible factors, and , you can write those in the parenthesis: .
 Then, use process of elimination to plug in the factors of 4 to find a combination that produces 11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is 4.^{[3] X Research source }
 By trial and error, try out this combination of factors . When you multiply them out, you get . If you combine the terms and , you get , which is the middle term you were aiming for. You have just factored the quadratic equation.
 As an example of trial and error, let's try checking a factoring combination for that is an error (does not work): = . If you combine those terms, you get . Though the factors 2 and 2 do multiply to make 4, the middle term does not work, because you needed to get , not .
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3Set each set of parenthesis equal to zero as separate equations. This will lead you to find two values for that will make the entire equation equal to zero, = 0. Now that you've factored the equation, all you have to do is put the expression in each set of parenthesis equal to zero. But why?  because to get zero by multiplying, we have the "principle, rule or property" that one factor must be zero, then at least one of the factors in parentheses, as must be zero; so, either (3x + 1) or else (x  4) must equal zero. So, you would write and also.

4Solve each "zeroed" equation independently. In a quadratic equation, there will be two possible values for x. Find x for each possible value of x one by one by isolating the variable and writing down the two solutions for x as the final solution. Here's how you do it:
 Solve 3x + 1 = 0
 3x = 1 ..... by subtracting
 3x/3 = 1/3 ..... by dividing
 x = 1/3 ..... simplified
 Solve x  4 = 0
 x = 4 ..... by subtracting
 x = (1/3, 4) ..... by making a set of possible, separate solutions, meaning x = 1/3, or x = 4 seem good.
 Solve 3x + 1 = 0

5Check x = 1/3 in (3x + 1)(x – 4) = 0:
We have (3[1/3] + 1)([1/3] – 4) ?=? 0 ..... by substituting (1 + 1)(4 1/3) ?=? 0 ..... by simplifying (0)(4 1/3) = 0 ..... by multiplying therefore 0 = 0 ..... Yes, x = 1/3 works 
6Check x = 4 in (3x + 1)(x  4) = 0:
We have (3[4] + 1)([4] – 4) ?=? 0 ..... by substituting (13)(4 – 4) ?=? 0 ..... by simplifying (13)(0) = 0 ..... by multiplying 0 = 0 ..... Yes, x = 4 works So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.
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Method 2
Method 2 of 3:Using the Quadratic Formula

1Combine all of the like terms and move them to one side of the equation. Move all of the terms to one side of the equal sign, keeping the term positive. Write the terms in descending order of degrees, so that the term comes first, followed by the term and the constant term.^{[4] X Research source } Here's how you do it:
 4x^{2}  5x  13 = x^{2} 5
 4x^{2}  x^{2}  5x  13 +5 = 0
 3x^{2}  5x  8 = 0

2Write down the quadratic formula. The quadratic formula is: ^{[5] X Research source }

3Identify the values of a, b, and c in the quadratic equation. The variable a is the coefficient of the x^{2} term, b is the coefficient of the x term, and c is the constant. For the equation 3x^{2} 5x  8 = 0, a = 3, b = 5, and c = 8. Write this down.

4Substitute the values of a, b, and c into the equation. Now that you know the values of the three variables, you can just plug them into the equation like this:
 {b +/√ (b^{2}  4ac)}/2
 {(5) +/√ ((5)^{2}  4(3)(8))}/2(3) =
 {(5) +/√ ((5)^{2}  (96))}/2(3)

5Do the math. After you've plugged in the numbers, do the remaining math to simplify positive or negative signs, multiply, or square the remaining terms. Here's how you do it:
 {(5) +/√ ((5)^{2}  (96))}/2(3) =
 {5 +/√(25 + 96)}/6
 {5 +/√(121)}/6

6Simplify the square root. If the number under the radical symbol is a perfect square, you will get a whole number. If the number is not a perfect square, then simplify to its simplest radical version. If the number is negative, and you're sure it's supposed to be negative, then the roots will be complex. In this example, √(121) = 11. You can write that x = (5 +/ 11)/6.

7Solve for the positive and negative answers. If you've eliminated the square root symbol, then you can keep going until you've found the positive and negative results for x. Now that you have (5 +/ 11)/6, you can write two options:
 (5 + 11)/6
 (5  11)/6

8Solve for the positive and negative answers. Just do the math:
 (5 + 11)/6 = 16/6
 (511)/6 = 6/6

9Simplify. To simplify each answer, just divide them by the largest number that is evenly divisible into both numbers. Divide the first fraction by 2, and divide the second by 6, and you have solved for x.
 16/6 = 8/3
 6/6 = 1
 x = (1, 8/3)
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Method 3
Method 3 of 3:
Completing the Square

1Move all of the terms to one side of the equation. Make sure that the a or x^{2} term is positive. Here's how you do it:^{[6] X Research source }
 2x^{2}  9 = 12x =
 2x^{2}  12x  9 = 0
 In this equation, the a term is 2, the b term is 12, and the c term is 9.

2Move the c term or constant to the other side. The constant term is the numerical term without a variable. Move it to the right side of the equation:
 2x^{2}  12x  9 = 0
 2x^{2}  12x = 9

3Divide both sides by the coefficient of the a or x^{2} term. If x^{2} has no term in front of it, and just has a coefficient of 1, then you can skip this step. In this case, you'll have to divide all of the terms by 2, like so:
 2x^{2}/2  12x/2 = 9/2 =
 x^{2}  6x = 9/2

4Divide b by two, square it, and add the result to both sides. The b term in this example is 6. Here's how you do it:
 6/2 = 3 =
 (3)^{2} = 9 =
 x^{2}  6x + 9 = 9/2 + 9

5Simplify both sides. Factor the terms on the left side to get (x3)(x3), or (x3)^{2}. Add the terms on the right side to get 9/2 + 9, or 9/2 + 18/2, which adds up to 27/2.

6Find the square root of both sides. The square root of (x3)^{2} is simply (x3). You can write the square root of 27/2 as ±√(27/2). Therefore, x  3 = ±√(27/2).

7Simplify the radical and solve for x. To simplify ±√(27/2), look for a perfect square within the numbers 27 or 2 or in their factors. The perfect square 9 can be found in 27, because 9 x 3 = 27. To take 9 out of the radical sign, pull out the number 9 from the radical, and write the number 3, its square root, outside the radical sign. Leave 3 in the numerator of the fraction under the radical sign, since that factor of 27 cannot be taken out, and leave 2 on the bottom. Then, move the constant 3 on the left side of the equation to the right, and write down your two solutions for x:
 x = 3 + 3(√6)/2
 x = 3  3(√6)/2)
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Community Q&A

QuestionI am totally lost. Are than any really simple shortcuts for one to use?DonaganTop AnswererSorry, no. Factoring quadratics is definitely a challenge for many people. Console yourself with the knowledge that you're not likely ever to need this skill in real life. Passing your next math test is another matter.

QuestionHow do I solve the following equation: Find the values of k for which the roots of the equation (k+4)x^2+(k+1)x+1=0 are real and equal?Community AnswerHaving a double root means that your espression is a perfect square. It also means that the discriminate is zero. Using either of these will give an equation in k that you can solve for the answer. A square takes the form (ax+b)^2= a^2 x^2 + 2abx + b^2. Obviously, then b = +1 or 1, leaving a=+/ (k+1)/2 = sqrt(k+4). Setting the discriminant of the original quadratic to zero gives (k+1)^2  4 (k+4) = 0. Whichever way you prefer you should get k=3 (x^2 2x +1 = (x1)^2) and k=5 (9x^2 + 6x +1 = (3x+1)^2).

QuestionWhat happened to the 6x in step 5?DonaganTop Answererx²6x+9 was factored into (x3)(x3) or (x3)².

QuestionHow can I solve the equation y^235y+306=0?Community AnswerFactor into (y18)(y17). The solutions are y = 17 and y = 18.

QuestionHow many square roots does a number have?OrangejewsCommunity AnswerPositive numbers have two square roots  the negative and positive version of each number. 0 only has one, 0. Negative numbers don't have any real square roots, but do have two complex square roots.

QuestionHow do I solve quadratic equations using the sum and product as the roots of the equation?Community Answera*x^2 [squared] + b*x = c = 0 is the general form for any quadratic equation. If the roots of this equation are k and m [suppose], then k + m = (b/a). & k*m = (c/a). Use these equations to solve for k and m or for x.

QuestionHow do I solve a quadratic equation that is in fractional form?DonaganTop AnswererRemove the fraction(s) by multiplying both sides of the equation by the denominator(s) of the fraction(s). Then proceed as usual.

QuestionHow do I factor x^2+5x4?Community AnswerYou cannot easily factor this expression because there does not exist two numbers whose sum is 5 and whose product is 4. We can confirm this by using the quadratic formula, where we find that x = 5/2 + sqrt(41)/2, which are irrational numbers.

QuestionHow do I solve the equation 4x100>6x+252?Community Answer176 +100 to each side to get rid of it. 4x>6x+352. 6x each side. 2x>352. Divide each side by 2. x>176. Done.

QuestionWhat is the process in which 6x is factored out in step 5?Prem ShahCommunity AnswerThe process starts when x^26x+9 gets factored out to (x3)(x3). Then, you use the FOIL method to multiply x*x, then 3x3x to get 6x. Then, you multiply 3 by 3 to get 9.
Tips
 As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plusorminus. Instead, we divide out any common factors  but ONLY if the factor is common to both of the constants AND the radical's coefficient.Thanks!
 If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example:^{[7] X Research source }Thanks!
 If the "b" is an even number, the formula is : {(b/2) +/ √(b/2)ac}/a.Thanks!
References
 ↑ https://www.mathsisfun.com/definitions/quadraticequation.html
 ↑ http://www.mathsisfun.com/algebra/factoringquadratics.html
 ↑ https://www.mathportal.org/algebra/solvingsystemoflinearequations/eliminationmethod.php
 ↑ https://www.purplemath.com/modules/solvquad4.htm
 ↑ http://www.purplemath.com/modules/quadform.htm
 ↑ http://www.mathsisfun.com/algebra/completingsquare.html
 ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm
About This Article
To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down!