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2x^{2}-2x-12-28=0
Subtract 28 from both sides.
2x^{2}-2x-40=0
Subtract 28 from -12 to get -40.
x^{2}-x-20=0
Divide both sides by 2.
a+b=-1 ab=1\left(-20\right)=-20
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.
1,-20 2,-10 4,-5
Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.
1-20=-19 2-10=-8 4-5=-1
Calculate the sum for each pair.
a=-5 b=4
The solution is the pair that gives sum -1.
\left(x^{2}-5x\right)+\left(4x-20\right)
Rewrite x^{2}-x-20 as \left(x^{2}-5x\right)+\left(4x-20\right).
x\left(x-5\right)+4\left(x-5\right)
Factor out x in the first and 4 in the second group.
\left(x-5\right)\left(x+4\right)
Factor out common term x-5 by using distributive property.
x=5 x=-4
To find equation solutions, solve x-5=0 and x+4=0.
2x^{2}-2x-12=28
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
2x^{2}-2x-12-28=28-28
Subtract 28 from both sides of the equation.
2x^{2}-2x-12-28=0
Subtracting 28 from itself leaves 0.
2x^{2}-2x-40=0
Subtract 28 from -12.
x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 2\left(-40\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -2 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-40\right)}}{2\times 2}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-8\left(-40\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 2}
Multiply -8 times -40.
x=\frac{-\left(-2\right)±\sqrt{324}}{2\times 2}
Add 4 to 320.
x=\frac{-\left(-2\right)±18}{2\times 2}
Take the square root of 324.
x=\frac{2±18}{2\times 2}
The opposite of -2 is 2.
x=\frac{2±18}{4}
Multiply 2 times 2.
x=\frac{20}{4}
Now solve the equation x=\frac{2±18}{4} when ± is plus. Add 2 to 18.
x=5
Divide 20 by 4.
x=\frac{-16}{4}
Now solve the equation x=\frac{2±18}{4} when ± is minus. Subtract 18 from 2.
x=-4
Divide -16 by 4.
x=5 x=-4
The equation is now solved.
2x^{2}-2x-12=28
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
2x^{2}-2x-12-\left(-12\right)=28-\left(-12\right)
Add 12 to both sides of the equation.
2x^{2}-2x=28-\left(-12\right)
Subtracting -12 from itself leaves 0.
2x^{2}-2x=40
Subtract -12 from 28.
\frac{2x^{2}-2x}{2}=\frac{40}{2}
Divide both sides by 2.
x^{2}+\frac{-2}{2}x=\frac{40}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-x=\frac{40}{2}
Divide -2 by 2.
x^{2}-x=20
Divide 40 by 2.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=20+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=20+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{81}{4}
Add 20 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=\frac{81}{4}
Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{9}{2} x-\frac{1}{2}=-\frac{9}{2}
Simplify.
x=5 x=-4
Add \frac{1}{2} to both sides of the equation.