3z(4z-6)=0

Solution for 3z(4z-6)=0 equation:

3z(4z-6)=0

We multiply parentheses

12z^2-18z=0

a = 12; b = -18; c = 0;
Δ = b2-4ac
Δ = -182-4·12·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-18}{2*12}=\frac{0}{24}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+18}{2*12}=\frac{36}{24}
=1+1/2
$