Solve 11x+12=x^2

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

Solve for x

$$x=-1$$

$$x=12$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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.

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

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 the first and -1 in the second group.

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

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 x^{2} from both sides.

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

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.

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

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}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-11±\sqrt{121+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-11±\sqrt{121+48}}{2\left(-1\right)}$$

Add 121 to 48.

$$x=\frac{-11±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-11±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-11±13}{-2}$$

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

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

Divide 2 by -2.

$$x=-1$$

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

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

Divide -24 by -2.

$$x=12$$

The equation is now solved.

$$x=-1 x=12$$

Steps for Completing the Square

Subtract x^{2} from both sides.

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

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}+11x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{11}{-1}x=-\frac{12}{-1}$$

Divide 11 by -1.

$$x^{2}-11x=-\frac{12}{-1}$$

Divide -12 by -1.

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

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.

$$x=12 x=-1$$