Solve -y^2=9y

$$- y ^ { 2 } = 9 y$$

Solve for y

$$y=-9$$

$$y=0$$

Steps Using the Quadratic Formula

Subtract 9y from both sides.

$$-y^{2}-9y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -9 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}}}{2\left(-1\right)}$$

Take the square root of \left(-9\right)^{2}.

$$y=\frac{-\left(-9\right)±9}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$y=\frac{9±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{9±9}{-2}$$

Now solve the equation y=\frac{9±9}{-2} when ± is plus. Add 9 to 9.

$$y=\frac{18}{-2}$$

Divide 18 by -2.

$$y=-9$$

Now solve the equation y=\frac{9±9}{-2} when ± is minus. Subtract 9 from 9.

$$y=\frac{0}{-2}$$

Divide 0 by -2.

$$y=0$$

The equation is now solved.

$$y=-9 y=0$$

Steps for Completing the Square

Subtract 9y from both sides.

$$-y^{2}-9y=0$$

Divide both sides by -1.

$$\frac{-y^{2}-9y}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\left(-\frac{9}{-1}\right)y=\frac{0}{-1}$$

Divide -9 by -1.

$$y^{2}+9y=\frac{0}{-1}$$

Divide 0 by -1.

$$y^{2}+9y=0$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+9y+\frac{81}{4}=\frac{81}{4}$$

Factor y^{2}+9y+\frac{81}{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}.

$$\left(y+\frac{9}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{9}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$y+\frac{9}{2}=\frac{9}{2} y+\frac{9}{2}=-\frac{9}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$y=0 y=-9$$