(3x+3)4x=132

Solution for (3x+3)4x=132 equation:

(3x+3)4x=132

We move all terms to the left:

(3x+3)4x-(132)=0

We multiply parentheses

12x^2+12x-132=0

a = 12; b = 12; c = -132;
Δ = b2-4ac
Δ = 122-4·12·(-132)
Δ = 6480
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:

$\sqrt{\Delta}=\sqrt{6480}=\sqrt{1296*5}=\sqrt{1296}*\sqrt{5}=36\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-36\sqrt{5}}{2*12}=\frac{-12-36\sqrt{5}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+36\sqrt{5}}{2*12}=\frac{-12+36\sqrt{5}}{24}
$