2^x*2^x+1=8

Solution for 2^x*2^x+1=8 equation:

2^x*2^x+1=8

We move all terms to the left:

2^x*2^x+1-(8)=0

We add all the numbers together, and all the variables

2^x*2^x-7=0

Wy multiply elements

4x^2-7=0

a = 4; b = 0; c = -7;
Δ = b2-4ac
Δ = 02-4·4·(-7)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*4}=\frac{0-4\sqrt{7}}{8}
=-\frac{4\sqrt{7}}{8}
=-\frac{\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*4}=\frac{0+4\sqrt{7}}{8}
=\frac{4\sqrt{7}}{8}
=\frac{\sqrt{7}}{2}
$