Solve -4y^2=12y

$$- 4 y ^ { 2 } = 12 y$$

Solve for y

$$y=-3$$

$$y=0$$

Steps Using Factoring

Subtract 12y from both sides.

$$-4y^{2}-12y=0$$

Factor out y.

$$y\left(-4y-12\right)=0$$

To find equation solutions, solve y=0 and -4y-12=0.

$$y=0 y=-3$$

Steps Using the Quadratic Formula

Subtract 12y from both sides.

$$-4y^{2}-12y=0$$

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

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\left(-4\right)}$$

Take the square root of \left(-12\right)^{2}.

$$y=\frac{-\left(-12\right)±12}{2\left(-4\right)}$$

The opposite of -12 is 12.

$$y=\frac{12±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$y=\frac{12±12}{-8}$$

Now solve the equation y=\frac{12±12}{-8} when ± is plus. Add 12 to 12.

$$y=\frac{24}{-8}$$

Divide 24 by -8.

$$y=-3$$

Now solve the equation y=\frac{12±12}{-8} when ± is minus. Subtract 12 from 12.

$$y=\frac{0}{-8}$$

Divide 0 by -8.

$$y=0$$

The equation is now solved.

$$y=-3 y=0$$

Steps for Completing the Square

Subtract 12y from both sides.

$$-4y^{2}-12y=0$$

Divide both sides by -4.

$$\frac{-4y^{2}-12y}{-4}=\frac{0}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y^{2}+\left(-\frac{12}{-4}\right)y=\frac{0}{-4}$$

Divide -12 by -4.

$$y^{2}+3y=\frac{0}{-4}$$

Divide 0 by -4.

$$y^{2}+3y=0$$

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+3y+\frac{9}{4}=\frac{9}{4}$$

Factor y^{2}+3y+\frac{9}{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}.

$$\left(y+\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$y+\frac{3}{2}=\frac{3}{2} y+\frac{3}{2}=-\frac{3}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$y=0 y=-3$$