2(4^2x+1)=128

Solution for 2(4^2x+1)=128 equation:

2(4^2x+1)=128

We move all terms to the left:

2(4^2x+1)-(128)=0

We multiply parentheses

8x^2+2-128=0

We add all the numbers together, and all the variables

8x^2-126=0

a = 8; b = 0; c = -126;
Δ = b2-4ac
Δ = 02-4·8·(-126)
Δ = 4032
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:

$\sqrt{\Delta}=\sqrt{4032}=\sqrt{576*7}=\sqrt{576}*\sqrt{7}=24\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{7}}{2*8}=\frac{0-24\sqrt{7}}{16}
=-\frac{24\sqrt{7}}{16}
=-\frac{3\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{7}}{2*8}=\frac{0+24\sqrt{7}}{16}
=\frac{24\sqrt{7}}{16}
=\frac{3\sqrt{7}}{2}
$