-3(y-5)(-y+4)=0

Solution for -3(y-5)(-y+4)=0 equation:

-3(y-5)(-y+4)=0

We add all the numbers together, and all the variables

-3(y-5)(-1y+4)=0

We multiply parentheses ..

-3(-1y^2+4y+5y-20)=0

We multiply parentheses

3y^2-12y-15y+60=0

We add all the numbers together, and all the variables

3y^2-27y+60=0

a = 3; b = -27; c = +60;
Δ = b2-4ac
Δ = -272-4·3·60
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3}{2*3}=\frac{24}{6}
=4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3}{2*3}=\frac{30}{6}
=5
$