Solve 2(0)^2+2y^2+12y=0

$$2 ( 0 ) ^ { 2 } + 2 y ^ { 2 } + 12 y = 0$$

Solve for y

$$y=-6$$

$$y=0$$

Steps Using Factoring

Calculate 0 to the power of 2 and get 0.

$$2\times 0+2y^{2}+12y=0$$

Multiply 2 and 0 to get 0.

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

Anything plus zero gives itself.

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

Factor out y.

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

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

$$y=0 y=-6$$

Steps Using the Quadratic Formula

Calculate 0 to the power of 2 and get 0.

$$2\times 0+2y^{2}+12y=0$$

Multiply 2 and 0 to get 0.

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

Anything plus zero gives itself.

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

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}.

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

Take the square root of 12^{2}.

$$y=\frac{-12±12}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{-12±12}{4}$$

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

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

Divide 0 by 4.

$$y=0$$

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

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

Divide -24 by 4.

$$y=-6$$

The equation is now solved.

$$y=0 y=-6$$

Steps for Completing the Square

Calculate 0 to the power of 2 and get 0.

$$2\times 0+2y^{2}+12y=0$$

Multiply 2 and 0 to get 0.

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

Anything plus zero gives itself.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 12 by 2.

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

Divide 0 by 2.

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

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.

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

Square 3.

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

Factor y^{2}+6y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y+3=3 y+3=-3$$

Subtract 3 from both sides of the equation.

$$y=0 y=-6$$