3z(z-4)=0

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

3z(z-4)=0

We multiply parentheses

3z^2-12z=0

a = 3; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·3·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*3}=\frac{0}{6}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*3}=\frac{24}{6}
=4
$