x^2-20x+100=8

Solution for x^2-20x+100=8 equation:

x^2-20x+100=8

We move all terms to the left:

x^2-20x+100-(8)=0

We add all the numbers together, and all the variables

x^2-20x+92=0

a = 1; b = -20; c = +92;
Δ = b2-4ac
Δ = -202-4·1·92
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{2}}{2*1}=\frac{20-4\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{2}}{2*1}=\frac{20+4\sqrt{2}}{2}
$