4(4^2x+4^-2x)=17

Solution for 4(4^2x+4^-2x)=17 equation:

4(4^2x+4^-2x)=17

We move all terms to the left:

4(4^2x+4^-2x)-(17)=0

We multiply parentheses

16x^2-8x-17=0

a = 16; b = -8; c = -17;
Δ = b2-4ac
Δ = -82-4·16·(-17)
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-24\sqrt{2}}{2*16}=\frac{8-24\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+24\sqrt{2}}{2*16}=\frac{8+24\sqrt{2}}{32}
$