2(2^2x+3)=32

Solution for 2(2^2x+3)=32 equation:

2(2^2x+3)=32

We move all terms to the left:

2(2^2x+3)-(32)=0

We multiply parentheses

4x^2+6-32=0

We add all the numbers together, and all the variables

4x^2-26=0

a = 4; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·4·(-26)
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{26}}{2*4}=\frac{0-4\sqrt{26}}{8}
=-\frac{4\sqrt{26}}{8}
=-\frac{\sqrt{26}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{26}}{2*4}=\frac{0+4\sqrt{26}}{8}
=\frac{4\sqrt{26}}{8}
=\frac{\sqrt{26}}{2}
$