Solve x^2=-7x-12

$$x ^ { 2 } = - 7 x - 12$$

Solve for x

$$x=-4$$

$$x=-3$$

Steps Using Factoring

Add 7x to both sides.

$$x^{2}+7x=-12$$

Add 12 to both sides.

$$x^{2}+7x+12=0$$

To solve the equation, factor x^{2}+7x+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.

$$a+b=7 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

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

$$\left(x+3\right)\left(x+4\right)$$

To find equation solutions, solve x+3=0 and x+4=0.

$$x=-3 x=-4$$

Steps Using Factoring By Grouping

Add 7x to both sides.

$$x^{2}+7x=-12$$

Add 12 to both sides.

$$x^{2}+7x+12=0$$

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.

$$a+b=7 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

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

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite x^{2}+7x+12 as \left(x^{2}+3x\right)+\left(4x+12\right).

$$\left(x^{2}+3x\right)+\left(4x+12\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(x+3\right)+4\left(x+3\right)$$

Factor out common term x+3 by using distributive property.

$$\left(x+3\right)\left(x+4\right)$$

To find equation solutions, solve x+3=0 and x+4=0.

$$x=-3 x=-4$$

Steps Using the Quadratic Formula

Add 7x to both sides.

$$x^{2}+7x=-12$$

Add 12 to both sides.

$$x^{2}+7x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 7 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 12}}{2}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 12}}{2}$$

Multiply -4 times 12.

$$x=\frac{-7±\sqrt{49-48}}{2}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-7±1}{2}$$

Now solve the equation x=\frac{-7±1}{2} when ± is plus. Add -7 to 1.

$$x=-\frac{6}{2}$$

Divide -6 by 2.

$$x=-3$$

Now solve the equation x=\frac{-7±1}{2} when ± is minus. Subtract 1 from -7.

$$x=-\frac{8}{2}$$

Divide -8 by 2.

$$x=-4$$

The equation is now solved.

$$x=-3 x=-4$$

Steps for Completing the Square

Add 7x to both sides.

$$x^{2}+7x=-12$$

Divide 7, the coefficient of the x term, by 2 to get \frac{7}{2}. Then add the square of \frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=-12+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=-12+\frac{49}{4}$$

Add -12 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{1}{4}$$

Factor x^{2}+7x+\frac{49}{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}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{1}{2} x+\frac{7}{2}=-\frac{1}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=-3 x=-4$$