4(3^2x-4)=36

Solution for 4(3^2x-4)=36 equation:

4(3^2x-4)=36

We move all terms to the left:

4(3^2x-4)-(36)=0

We multiply parentheses

12x^2-16-36=0

We add all the numbers together, and all the variables

12x^2-52=0

a = 12; b = 0; c = -52;
Δ = b2-4ac
Δ = 02-4·12·(-52)
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{39}}{2*12}=\frac{0-8\sqrt{39}}{24}
=-\frac{8\sqrt{39}}{24}
=-\frac{\sqrt{39}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{39}}{2*12}=\frac{0+8\sqrt{39}}{24}
=\frac{8\sqrt{39}}{24}
=\frac{\sqrt{39}}{3}
$