Solve 2x^2=-12x-18

$$2 x ^ { 2 } = - 12 x - 18$$

Solve for x

$$x=-3$$

Steps Using Factoring By Grouping

Add 12x to both sides.

$$2x^{2}+12x=-18$$

Add 18 to both sides.

$$2x^{2}+12x+18=0$$

Divide both sides by 2.

$$x^{2}+6x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\times 9=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 6.

$$a=3 b=3$$

Rewrite x^{2}+6x+9 as \left(x^{2}+3x\right)+\left(3x+9\right).

$$\left(x^{2}+3x\right)+\left(3x+9\right)$$

Factor out x in the first and 3 in the second group.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve x+3=0.

$$x=-3$$

Steps Using the Quadratic Formula

Add 12x to both sides.

$$2x^{2}+12x=-18$$

Add 18 to both sides.

$$2x^{2}+12x+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 12 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 2\times 18}}{2\times 2}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 2\times 18}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-12±\sqrt{144-8\times 18}}{2\times 2}$$

Multiply -8 times 18.

$$x=\frac{-12±\sqrt{144-144}}{2\times 2}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\times 2}$$

Take the square root of 0.

$$x=-\frac{12}{2\times 2}$$

Multiply 2 times 2.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

Steps for Completing the Square

Add 12x to both sides.

$$2x^{2}+12x=-18$$

Divide both sides by 2.

$$\frac{2x^{2}+12x}{2}=-\frac{18}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{12}{2}x=-\frac{18}{2}$$

Divide 12 by 2.

$$x^{2}+6x=-\frac{18}{2}$$

Divide -18 by 2.

$$x^{2}+6x=-9$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-9+3^{2}$$

Square 3.

$$x^{2}+6x+9=-9+9$$

Add -9 to 9.

$$x^{2}+6x+9=0$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x+3=0 x+3=0$$

Subtract 3 from both sides of the equation.

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

The equation is now solved. Solutions are the same.

$$x=-3$$