(6x-12)(3x+3)=180

Solution for (6x-12)(3x+3)=180 equation:

(6x-12)(3x+3)=180

We move all terms to the left:

(6x-12)(3x+3)-(180)=0

We multiply parentheses ..

(+18x^2+18x-36x-36)-180=0

We get rid of parentheses

18x^2+18x-36x-36-180=0

We add all the numbers together, and all the variables

18x^2-18x-216=0

a = 18; b = -18; c = -216;
Δ = b2-4ac
Δ = -182-4·18·(-216)
Δ = 15876
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15876}=126$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-126}{2*18}=\frac{-108}{36}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+126}{2*18}=\frac{144}{36}
=4
$