-4(z+2)(-z-3)=0

Solution for -4(z+2)(-z-3)=0 equation:

-4(z+2)(-z-3)=0

We add all the numbers together, and all the variables

-4(z+2)(-1z-3)=0

We multiply parentheses ..

-4(-1z^2-3z-2z-6)=0

We multiply parentheses

4z^2+12z+8z+24=0

We add all the numbers together, and all the variables

4z^2+20z+24=0

a = 4; b = 20; c = +24;
Δ = b2-4ac
Δ = 202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*4}=\frac{-24}{8}
=-3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*4}=\frac{-16}{8}
=-2
$