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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-3(-1z^2+2z+4z-8)=0

We multiply parentheses

3z^2-6z-12z+24=0

We add all the numbers together, and all the variables

3z^2-18z+24=0

a = 3; b = -18; c = +24;
Δ = b2-4ac
Δ = -182-4·3·24
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*3}=\frac{12}{6}
=2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*3}=\frac{24}{6}
=4
$