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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We multiply parentheses

-4z^2-16z+20z+80=0

We add all the numbers together, and all the variables

-4z^2+4z+80=0

a = -4; b = 4; c = +80;
Δ = b2-4ac
Δ = 42-4·(-4)·80
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*-4}=\frac{-40}{-8}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*-4}=\frac{32}{-8}
=-4
$