(y+5)(8y-32)=0

Solution for (y+5)(8y-32)=0 equation:

(y+5)(8y-32)=0

We multiply parentheses ..

(+8y^2-32y+40y-160)=0

We get rid of parentheses

8y^2-32y+40y-160=0

We add all the numbers together, and all the variables

8y^2+8y-160=0

a = 8; b = 8; c = -160;
Δ = b2-4ac
Δ = 82-4·8·(-160)
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*8}=\frac{-80}{16}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*8}=\frac{64}{16}
=4
$