4(3^2x+1)+5=329

Solution for 4(3^2x+1)+5=329 equation:

4(3^2x+1)+5=329

We move all terms to the left:

4(3^2x+1)+5-(329)=0

We add all the numbers together, and all the variables

4(3^2x+1)-324=0

We multiply parentheses

12x^2+4-324=0

We add all the numbers together, and all the variables

12x^2-320=0

a = 12; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·12·(-320)
Δ = 15360
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:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*12}=\frac{0-32\sqrt{15}}{24}
=-\frac{32\sqrt{15}}{24}
=-\frac{4\sqrt{15}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*12}=\frac{0+32\sqrt{15}}{24}
=\frac{32\sqrt{15}}{24}
=\frac{4\sqrt{15}}{3}
$