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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-4(-1z^2+3z+5z-15)=0

We multiply parentheses

4z^2-12z-20z+60=0

We add all the numbers together, and all the variables

4z^2-32z+60=0

a = 4; b = -32; c = +60;
Δ = b2-4ac
Δ = -322-4·4·60
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8}{2*4}=\frac{24}{8}
=3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8}{2*4}=\frac{40}{8}
=5
$