(x+12)(x+6)=4x^2+36x

Solution for (x+12)(x+6)=4x^2+36x equation:

(x+12)(x+6)=4x^2+36x

We move all terms to the left:

(x+12)(x+6)-(4x^2+36x)=0

We get rid of parentheses

-4x^2+(x+12)(x+6)-36x=0

We multiply parentheses ..

-4x^2+(+x^2+6x+12x+72)-36x=0

We get rid of parentheses

-4x^2+x^2+6x+12x-36x+72=0

We add all the numbers together, and all the variables

-3x^2-18x+72=0

a = -3; b = -18; c = +72;
Δ = b2-4ac
Δ = -182-4·(-3)·72
Δ = 1188
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:

$\sqrt{\Delta}=\sqrt{1188}=\sqrt{36*33}=\sqrt{36}*\sqrt{33}=6\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{33}}{2*-3}=\frac{18-6\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{33}}{2*-3}=\frac{18+6\sqrt{33}}{-6}
$