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

Solution for -4(z+1)(-z-5)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We multiply parentheses

4z^2+20z+4z+20=0

We add all the numbers together, and all the variables

4z^2+24z+20=0

a = 4; b = 24; c = +20;
Δ = b2-4ac
Δ = 242-4·4·20
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*4}=\frac{-40}{8}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*4}=\frac{-8}{8}
=-1
$