What are the solutions to the equation 0=(2y-1)(8-y) ?

The solutions to the equation 0=(2y-1)(8-y) are y= 1/2 ,y=8

Find the zeros of 0=(2y-1)(8-y)

The zeros of 0=(2y-1)(8-y) are y= 1/2 ,y=8

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

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

Solution steps

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

Switch sides

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 y - 1 = 0 or 8 - y = 0

Solve

2 y - 1 = 0 : y = \frac{1}{2}

Solve

8 - y = 0 : y = 8

The solutions to the quadratic equation are:

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

216 x^{3} - 27

\frac{( x - 7 )}{( 4 x - 15 )} > 8

- 16 = \frac{8 x}{5}

3 n^{2} + 8 n + 16 = 2 n^{2}

2 \geq \frac{n}{4} - 2

What are the solutions to the equation 0=(2y-1)(8-y) ?
The solutions to the equation 0=(2y-1)(8-y) are y= 1/2 ,y=8
Find the zeros of 0=(2y-1)(8-y)
The zeros of 0=(2y-1)(8-y) are y= 1/2 ,y=8