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

Solution for (4-y)(2y+8)=0 equation:

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

We add all the numbers together, and all the variables

(-1y+4)(2y+8)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2y^2+32=0

a = -2; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-2)·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-2}=\frac{-16}{-4}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-2}=\frac{16}{-4}
=-4
$