Solve for x

x=-4

x=5

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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.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

Linear equation

y = 3x + 4

Arithmetic

699 * 533

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}