2(4^2x+2)=128

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

2(4^2x+2)=128

We move all terms to the left:

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

We multiply parentheses

8x^2+4-128=0

We add all the numbers together, and all the variables

8x^2-124=0

a = 8; b = 0; c = -124;
Δ = b2-4ac
Δ = 02-4·8·(-124)
Δ = 3968
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:

$\sqrt{\Delta}=\sqrt{3968}=\sqrt{64*62}=\sqrt{64}*\sqrt{62}=8\sqrt{62}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{62}}{2*8}=\frac{0-8\sqrt{62}}{16}
=-\frac{8\sqrt{62}}{16}
=-\frac{\sqrt{62}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{62}}{2*8}=\frac{0+8\sqrt{62}}{16}
=\frac{8\sqrt{62}}{16}
=\frac{\sqrt{62}}{2}
$