(3x-3)(x+2)=72

Solution for (3x-3)(x+2)=72 equation:

(3x-3)(x+2)=72

We move all terms to the left:

(3x-3)(x+2)-(72)=0

We multiply parentheses ..

(+3x^2+6x-3x-6)-72=0

We get rid of parentheses

3x^2+6x-3x-6-72=0

We add all the numbers together, and all the variables

3x^2+3x-78=0

a = 3; b = 3; c = -78;
Δ = b2-4ac
Δ = 32-4·3·(-78)
Δ = 945
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{105}}{2*3}=\frac{-3-3\sqrt{105}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{105}}{2*3}=\frac{-3+3\sqrt{105}}{6}
$