Solve 3(z+2)^2=12

$$3 ( z + 2 ) ^ { 2 } = 12$$

Solve for z

$$z=0$$

$$z=-4$$

Steps Using Factoring

Divide both sides by 3.

$$\left(z+2\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(z+2\right)^{2}=4$$

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

$$z^{2}+4z+4=4$$

Subtract 4 from both sides.

$$z^{2}+4z+4-4=0$$

Subtract 4 from 4 to get 0.

$$z^{2}+4z=0$$

Factor out z.

$$z\left(z+4\right)=0$$

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

$$z=0 z=-4$$

Steps Using the Quadratic Formula

Divide both sides by 3.

$$\left(z+2\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(z+2\right)^{2}=4$$

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

$$z^{2}+4z+4=4$$

Subtract 4 from both sides.

$$z^{2}+4z+4-4=0$$

Subtract 4 from 4 to get 0.

$$z^{2}+4z=0$$

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

$$z=\frac{-4±\sqrt{4^{2}}}{2}$$

Take the square root of 4^{2}.

$$z=\frac{-4±4}{2}$$

Now solve the equation z=\frac{-4±4}{2} when ± is plus. Add -4 to 4.

$$z=\frac{0}{2}$$

Divide 0 by 2.

$$z=0$$

Now solve the equation z=\frac{-4±4}{2} when ± is minus. Subtract 4 from -4.

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

Divide -8 by 2.

$$z=-4$$

The equation is now solved.

$$z=0 z=-4$$

Steps for Completing the Square

Divide both sides by 3.

$$\left(z+2\right)^{2}=\frac{12}{3}$$

Divide 12 by 3 to get 4.

$$\left(z+2\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(z+2\right)^{2}}=\sqrt{4}$$

Simplify.

$$z+2=2 z+2=-2$$

Subtract 2 from both sides of the equation.

$$z=0 z=-4$$