4(3^2x+1)=64.

Solution for 4(3^2x+1)=64. equation:

4(3^2x+1)=64.

We move all terms to the left:

4(3^2x+1)-(64.)=0

We add all the numbers together, and all the variables

4(3^2x+1)-64=0

We multiply parentheses

12x^2+4-64=0

We add all the numbers together, and all the variables

12x^2-60=0

a = 12; b = 0; c = -60;
Δ = b2-4ac
Δ = 02-4·12·(-60)
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{5}}{2*12}=\frac{0-24\sqrt{5}}{24}
=-\frac{24\sqrt{5}}{24}
=-\sqrt{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{5}}{2*12}=\frac{0+24\sqrt{5}}{24}
=\frac{24\sqrt{5}}{24}
=\sqrt{5}
$