(3x+x)3x=72

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

(3x+x)3x=72

We move all terms to the left:

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

We add all the numbers together, and all the variables

(+4x)3x-72=0

We multiply parentheses

12x^2-72=0

a = 12; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·12·(-72)
Δ = 3456
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:

$\sqrt{\Delta}=\sqrt{3456}=\sqrt{576*6}=\sqrt{576}*\sqrt{6}=24\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{6}}{2*12}=\frac{0-24\sqrt{6}}{24}
=-\frac{24\sqrt{6}}{24}
=-\sqrt{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{6}}{2*12}=\frac{0+24\sqrt{6}}{24}
=\frac{24\sqrt{6}}{24}
=\sqrt{6}
$