Solve 2y^2-1,5y-5/y+2=2y-5/2

$$\frac { 2 y ^ { 2 } - 1,5 y - 5 } { y + 2 } = \frac { 2 y - 5 } { 2 }$$

Solve for y

$$y=1$$

$$y=0$$

Steps Using Factoring

Variable y cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 2\left(y+2\right), the least common multiple of y+2;2.

$$2\left(2y^{2}-1,5y-5\right)=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply 2 by 2y^{2}-1,5y-5.

$$4y^{2}-3y-10=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply y+2 by 2y-5 and combine like terms.

$$4y^{2}-3y-10=2y^{2}-y-10$$

Subtract 2y^{2} from both sides.

$$4y^{2}-3y-10-2y^{2}=-y-10$$

Combine 4y^{2} and -2y^{2} to get 2y^{2}.

$$2y^{2}-3y-10=-y-10$$

Add y to both sides.

$$2y^{2}-3y-10+y=-10$$

Combine -3y and y to get -2y.

$$2y^{2}-2y-10=-10$$

Add 10 to both sides.

$$2y^{2}-2y-10+10=0$$

Add -10 and 10 to get 0.

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

Factor out y.

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

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

$$y=0 y=1$$

Steps Using the Quadratic Formula

Variable y cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 2\left(y+2\right), the least common multiple of y+2;2.

$$2\left(2y^{2}-1,5y-5\right)=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply 2 by 2y^{2}-1,5y-5.

$$4y^{2}-3y-10=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply y+2 by 2y-5 and combine like terms.

$$4y^{2}-3y-10=2y^{2}-y-10$$

Subtract 2y^{2} from both sides.

$$4y^{2}-3y-10-2y^{2}=-y-10$$

Combine 4y^{2} and -2y^{2} to get 2y^{2}.

$$2y^{2}-3y-10=-y-10$$

Add y to both sides.

$$2y^{2}-3y-10+y=-10$$

Combine -3y and y to get -2y.

$$2y^{2}-2y-10=-10$$

Add 10 to both sides.

$$2y^{2}-2y-10+10=0$$

Add -10 and 10 to get 0.

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

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

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

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

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

The opposite of -2 is 2.

$$y=\frac{2±2}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{2±2}{4}$$

Now solve the equation y=\frac{2±2}{4} when ± is plus. Add 2 to 2.

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

Divide 4 by 4.

$$y=1$$

Now solve the equation y=\frac{2±2}{4} when ± is minus. Subtract 2 from 2.

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

Divide 0 by 4.

$$y=0$$

The equation is now solved.

$$y=1 y=0$$

Steps for Completing the Square

Variable y cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 2\left(y+2\right), the least common multiple of y+2;2.

$$2\left(2y^{2}-1,5y-5\right)=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply 2 by 2y^{2}-1,5y-5.

$$4y^{2}-3y-10=\left(y+2\right)\left(2y-5\right)$$

Use the distributive property to multiply y+2 by 2y-5 and combine like terms.

$$4y^{2}-3y-10=2y^{2}-y-10$$

Subtract 2y^{2} from both sides.

$$4y^{2}-3y-10-2y^{2}=-y-10$$

Combine 4y^{2} and -2y^{2} to get 2y^{2}.

$$2y^{2}-3y-10=-y-10$$

Add y to both sides.

$$2y^{2}-3y-10+y=-10$$

Combine -3y and y to get -2y.

$$2y^{2}-2y-10=-10$$

Add 10 to both sides.

$$2y^{2}-2y=-10+10$$

Add -10 and 10 to get 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

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

Divide 0 by 2.

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

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.

$$y=1 y=0$$