Solve x^2=11x+12

$$x ^ { 2 } = 11 x + 12$$

Solve for x

$$x=-1$$

$$x=12$$

Steps Using Factoring

Subtract 11x from both sides.

$$x^{2}-11x=12$$

Subtract 12 from both sides.

$$x^{2}-11x-12=0$$

To solve the equation, factor x^{2}-11x-12 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.

$$a+b=-11 ab=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

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

$$\left(x-12\right)\left(x+1\right)$$

To find equation solutions, solve x-12=0 and x+1=0.

$$x=12 x=-1$$

Steps Using Factoring By Grouping

Subtract 11x from both sides.

$$x^{2}-11x=12$$

Subtract 12 from both sides.

$$x^{2}-11x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\left(-12\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

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

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

Factor out x in x^{2}-12x.

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

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

$$\left(x-12\right)\left(x+1\right)$$

To find equation solutions, solve x-12=0 and x+1=0.

$$x=12 x=-1$$

Steps Using the Quadratic Formula

Subtract 11x from both sides.

$$x^{2}-11x=12$$

Subtract 12 from both sides.

$$x^{2}-11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-12\right)}}{2}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2}$$

Add 121 to 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$x=\frac{-\left(-11\right)±13}{2}$$

The opposite of -11 is 11.

$$x=\frac{11±13}{2}$$

Now solve the equation x=\frac{11±13}{2} when ± is plus. Add 11 to 13.

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

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{11±13}{2} when ± is minus. Subtract 13 from 11.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=12 x=-1$$

Steps for Completing the Square

Subtract 11x from both sides.

$$x^{2}-11x=12$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=12+\frac{121}{4}$$

Add 12 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{169}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{13}{2} x-\frac{11}{2}=-\frac{13}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=12 x=-1$$