(8y-32)(y+1)=0

Solution for (8y-32)(y+1)=0 equation:

(8y-32)(y+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8y^2-24y-32=0

a = 8; b = -24; c = -32;
Δ = b2-4ac
Δ = -242-4·8·(-32)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-40}{2*8}=\frac{-16}{16}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+40}{2*8}=\frac{64}{16}
=4
$