Solve x^2=1x+132

$$x ^ { 2 } = 1 x + 132$$

Solve for x

$$x=-11$$

$$x=12$$

Steps Using Factoring

Subtract 1x from both sides.

$$x^{2}-x=132$$

Subtract 132 from both sides.

$$x^{2}-x-132=0$$

To solve the equation, factor x^{2}-x-132 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-132$$

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 -132.

$$1,-132 2,-66 3,-44 4,-33 6,-22 11,-12$$

Calculate the sum for each pair.

$$1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1$$

The solution is the pair that gives sum -1.

$$a=-12 b=11$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-12\right)\left(x+11\right)$$

To find equation solutions, solve x-12=0 and x+11=0.

$$x=12 x=-11$$

Steps Using Factoring By Grouping

Subtract 1x from both sides.

$$x^{2}-x=132$$

Subtract 132 from both sides.

$$x^{2}-x-132=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-132. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-132\right)=-132$$

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 -132.

$$1,-132 2,-66 3,-44 4,-33 6,-22 11,-12$$

Calculate the sum for each pair.

$$1-132=-131 2-66=-64 3-44=-41 4-33=-29 6-22=-16 11-12=-1$$

The solution is the pair that gives sum -1.

$$a=-12 b=11$$

Rewrite x^{2}-x-132 as \left(x^{2}-12x\right)+\left(11x-132\right).

$$\left(x^{2}-12x\right)+\left(11x-132\right)$$

Factor out x in the first and 11 in the second group.

$$x\left(x-12\right)+11\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+11\right)$$

To find equation solutions, solve x-12=0 and x+11=0.

$$x=12 x=-11$$

Steps Using the Quadratic Formula

Subtract 1x from both sides.

$$x^{2}-x=132$$

Subtract 132 from both sides.

$$x^{2}-x-132=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -132 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-132\right)}}{2}$$

Multiply -4 times -132.

$$x=\frac{-\left(-1\right)±\sqrt{1+528}}{2}$$

Add 1 to 528.

$$x=\frac{-\left(-1\right)±\sqrt{529}}{2}$$

Take the square root of 529.

$$x=\frac{-\left(-1\right)±23}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±23}{2}$$

Now solve the equation x=\frac{1±23}{2} when ± is plus. Add 1 to 23.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{1±23}{2} when ± is minus. Subtract 23 from 1.

$$x=-\frac{22}{2}$$

Divide -22 by 2.

$$x=-11$$

The equation is now solved.

$$x=12 x=-11$$

Steps for Completing the Square

Subtract 1x from both sides.

$$x^{2}-x=132$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=132+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=132+\frac{1}{4}$$

Add 132 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{529}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{529}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{529}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{23}{2} x-\frac{1}{2}=-\frac{23}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=12 x=-11$$