2(4^2x-4)=106

Solution for 2(4^2x-4)=106 equation:

2(4^2x-4)=106

We move all terms to the left:

2(4^2x-4)-(106)=0

We multiply parentheses

8x^2-8-106=0

We add all the numbers together, and all the variables

8x^2-114=0

a = 8; b = 0; c = -114;
Δ = b2-4ac
Δ = 02-4·8·(-114)
Δ = 3648
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:

$\sqrt{\Delta}=\sqrt{3648}=\sqrt{64*57}=\sqrt{64}*\sqrt{57}=8\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{57}}{2*8}=\frac{0-8\sqrt{57}}{16}
=-\frac{8\sqrt{57}}{16}
=-\frac{\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{57}}{2*8}=\frac{0+8\sqrt{57}}{16}
=\frac{8\sqrt{57}}{16}
=\frac{\sqrt{57}}{2}
$