Solve 0=(2y-1)(8-y)

$$0 = ( 2 y - 1 ) ( 8 - y )$$

Solve for y

$$y=8$$

$$y=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2y-1 by 8-y and combine like terms.

$$0=17y-2y^{2}-8$$

Swap sides so that all variable terms are on the left hand side.

$$17y-2y^{2}-8=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2y^{2}+17y-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2y^{2}+ay+by-8. To find a and b, set up a system to be solved.

$$a+b=17 ab=-2\left(-8\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 17.

$$a=16 b=1$$

Rewrite -2y^{2}+17y-8 as \left(-2y^{2}+16y\right)+\left(y-8\right).

$$\left(-2y^{2}+16y\right)+\left(y-8\right)$$

Factor out 2y in the first and -1 in the second group.

$$2y\left(-y+8\right)-\left(-y+8\right)$$

Factor out common term -y+8 by using distributive property.

$$\left(-y+8\right)\left(2y-1\right)$$

To find equation solutions, solve -y+8=0 and 2y-1=0.

$$y=8 y=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2y-1 by 8-y and combine like terms.

$$0=17y-2y^{2}-8$$

Swap sides so that all variable terms are on the left hand side.

$$17y-2y^{2}-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2y^{2}+17y-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 17 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-17±\sqrt{17^{2}-4\left(-2\right)\left(-8\right)}}{2\left(-2\right)}$$

Square 17.

$$y=\frac{-17±\sqrt{289-4\left(-2\right)\left(-8\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$y=\frac{-17±\sqrt{289+8\left(-8\right)}}{2\left(-2\right)}$$

Multiply 8 times -8.

$$y=\frac{-17±\sqrt{289-64}}{2\left(-2\right)}$$

Add 289 to -64.

$$y=\frac{-17±\sqrt{225}}{2\left(-2\right)}$$

Take the square root of 225.

$$y=\frac{-17±15}{2\left(-2\right)}$$

Multiply 2 times -2.

$$y=\frac{-17±15}{-4}$$

Now solve the equation y=\frac{-17±15}{-4} when ± is plus. Add -17 to 15.

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

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-17±15}{-4} when ± is minus. Subtract 15 from -17.

$$y=-\frac{32}{-4}$$

Divide -32 by -4.

$$y=8$$

The equation is now solved.

$$y=\frac{1}{2} y=8$$

Steps for Completing the Square

Use the distributive property to multiply 2y-1 by 8-y and combine like terms.

$$0=17y-2y^{2}-8$$

Swap sides so that all variable terms are on the left hand side.

$$17y-2y^{2}-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$17y-2y^{2}=8$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2y^{2}+17y=8$$

Divide both sides by -2.

$$\frac{-2y^{2}+17y}{-2}=\frac{8}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$y^{2}+\frac{17}{-2}y=\frac{8}{-2}$$

Divide 17 by -2.

$$y^{2}-\frac{17}{2}y=\frac{8}{-2}$$

Divide 8 by -2.

$$y^{2}-\frac{17}{2}y=-4$$

Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{17}{2}y+\left(-\frac{17}{4}\right)^{2}=-4+\left(-\frac{17}{4}\right)^{2}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{17}{2}y+\frac{289}{16}=-4+\frac{289}{16}$$

Add -4 to \frac{289}{16}.

$$y^{2}-\frac{17}{2}y+\frac{289}{16}=\frac{225}{16}$$

Factor y^{2}-\frac{17}{2}y+\frac{289}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{17}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{17}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$y-\frac{17}{4}=\frac{15}{4} y-\frac{17}{4}=-\frac{15}{4}$$

Add \frac{17}{4} to both sides of the equation.

$$y=8 y=\frac{1}{2}$$