2x^2-20x+41=0

Solution for 2x^2-20x+41=0 equation:

2x^2-20x+41=0

a = 2; b = -20; c = +41;
Δ = b2-4ac
Δ = -202-4·2·41
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{2}}{2*2}=\frac{20-6\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{2}}{2*2}=\frac{20+6\sqrt{2}}{4}
$