Solve 2(x+3)^2-18=0

$$2 ( x + 3 ) ^ { 2 } - 18 = 0$$

Solve for x

$$x=0$$

$$x=-6$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2\left(x^{2}+6x+9\right)-18=0$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 18 from 18 to get 0.

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

Factor out x.

$$x\left(2x+12\right)=0$$

To find equation solutions, solve x=0 and 2x+12=0.

$$x=0 x=-6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2\left(x^{2}+6x+9\right)-18=0$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 18 from 18 to get 0.

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

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

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

Take the square root of 12^{2}.

$$x=\frac{-12±12}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-12±12}{4}$$

Now solve the equation x=\frac{-12±12}{4} when ± is plus. Add -12 to 12.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

Now solve the equation x=\frac{-12±12}{4} when ± is minus. Subtract 12 from -12.

$$x=-\frac{24}{4}$$

Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=0 x=-6$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2\left(x^{2}+6x+9\right)-18=0$$

Use the distributive property to multiply 2 by x^{2}+6x+9.

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

Subtract 18 from 18 to get 0.

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

Divide both sides by 2.

$$\frac{2x^{2}+12x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{12}{2}x=\frac{0}{2}$$

Divide 12 by 2.

$$x^{2}+6x=\frac{0}{2}$$

Divide 0 by 2.

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

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.

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

Square 3.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 3 from both sides of the equation.

$$x=0 x=-6$$