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

Solution for (2y+1)(8-y)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2y^2+16y-1y+8)=0

We get rid of parentheses

-2y^2+16y-1y+8=0

We add all the numbers together, and all the variables

-2y^2+15y+8=0

a = -2; b = 15; c = +8;
Δ = b2-4ac
Δ = 152-4·(-2)·8
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-17}{2*-2}=\frac{-32}{-4}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+17}{2*-2}=\frac{2}{-4}
=-1/2
$