X^2-20x+20=0

Solution for X^2-20x+20=0 equation:

X^2-20X+20=0

a = 1; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·1·20
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{5}}{2*1}=\frac{20-8\sqrt{5}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{5}}{2*1}=\frac{20+8\sqrt{5}}{2}
$