Solve y^2-y/y+3=0

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

Solve for y

$$y=1$$

$$y=0$$

Steps Using Factoring

Variable y cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by y+3.

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

Factor out y.

$$y\left(y-1\right)=0$$

To find equation solutions, solve y=0 and y-1=0.

$$y=0 y=1$$

Steps Using the Quadratic Formula

Variable y cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by y+3.

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

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

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

Take the square root of 1.

$$y=\frac{-\left(-1\right)±1}{2}$$

The opposite of -1 is 1.

$$y=\frac{1±1}{2}$$

Now solve the equation y=\frac{1±1}{2} when ± is plus. Add 1 to 1.

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

Divide 2 by 2.

$$y=1$$

Now solve the equation y=\frac{1±1}{2} when ± is minus. Subtract 1 from 1.

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

Divide 0 by 2.

$$y=0$$

The equation is now solved.

$$y=1 y=0$$

Steps for Completing the Square

Variable y cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by y+3.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-y+\frac{1}{4}=\frac{1}{4}$$

Factor y^{2}-y+\frac{1}{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}.

$$\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y-\frac{1}{2}=\frac{1}{2} y-\frac{1}{2}=-\frac{1}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$y=1 y=0$$