(6x+12)(3x-12)=78

Solution for (6x+12)(3x-12)=78 equation:

(6x+12)(3x-12)=78

We move all terms to the left:

(6x+12)(3x-12)-(78)=0

We multiply parentheses ..

(+18x^2-72x+36x-144)-78=0

We get rid of parentheses

18x^2-72x+36x-144-78=0

We add all the numbers together, and all the variables

18x^2-36x-222=0

a = 18; b = -36; c = -222;
Δ = b2-4ac
Δ = -362-4·18·(-222)
Δ = 17280
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:

$\sqrt{\Delta}=\sqrt{17280}=\sqrt{576*30}=\sqrt{576}*\sqrt{30}=24\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-24\sqrt{30}}{2*18}=\frac{36-24\sqrt{30}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+24\sqrt{30}}{2*18}=\frac{36+24\sqrt{30}}{36}
$