Solve 2y^2-3y-5y=0

$$2 { y }^{ 2 } -3y-5y=0$$

Solve for y

$$y=4$$

$$y=0$$

Steps Using Factoring

Factor out y.

$$y\left(2y-3-5\right)=0$$

To find equation solutions, solve y=0 and 2y-8=0.

$$y=0 y=4$$

Steps Using the Quadratic Formula

Combine -3y and -5y to get -8y.

$$2y^{2}-8y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 2}$$

Take the square root of \left(-8\right)^{2}.

$$y=\frac{-\left(-8\right)±8}{2\times 2}$$

The opposite of -8 is 8.

$$y=\frac{8±8}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{8±8}{4}$$

Now solve the equation y=\frac{8±8}{4} when ± is plus. Add 8 to 8.

$$y=\frac{16}{4}$$

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{8±8}{4} when ± is minus. Subtract 8 from 8.

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

Divide 0 by 4.

$$y=0$$

The equation is now solved.

$$y=4 y=0$$

Steps for Completing the Square

Combine -3y and -5y to get -8y.

$$2y^{2}-8y=0$$

Divide both sides by 2.

$$\frac{2y^{2}-8y}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\left(-\frac{8}{2}\right)y=\frac{0}{2}$$

Divide -8 by 2.

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

Divide 0 by 2.

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

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

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

Factor y^{2}-4y+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-2=2 y-2=-2$$

Add 2 to both sides of the equation.

$$y=4 y=0$$