(3z-3)(2z)=64

Solution for (3z-3)(2z)=64 equation:

(3z-3)(2z)=64

We move all terms to the left:

(3z-3)(2z)-(64)=0

We multiply parentheses

6z^2-6z-64=0

a = 6; b = -6; c = -64;
Δ = b2-4ac
Δ = -62-4·6·(-64)
Δ = 1572
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{393}}{2*6}=\frac{6-2\sqrt{393}}{12}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{393}}{2*6}=\frac{6+2\sqrt{393}}{12}
$