-3(y-8)(-y+4)=0

Solution for -3(y-8)(-y+4)=0 equation:

-3(y-8)(-y+4)=0

We add all the numbers together, and all the variables

-3(y-8)(-1y+4)=0

We multiply parentheses ..

-3(-1y^2+4y+8y-32)=0

We multiply parentheses

3y^2-12y-24y+96=0

We add all the numbers together, and all the variables

3y^2-36y+96=0

a = 3; b = -36; c = +96;
Δ = b2-4ac
Δ = -362-4·3·96
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*3}=\frac{24}{6}
=4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*3}=\frac{48}{6}
=8
$