x^2-17x+41=0

Solution for x^2-17x+41=0 equation:

x^2-17x+41=0

a = 1; b = -17; c = +41;
Δ = b2-4ac
Δ = -172-4·1·41
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-5\sqrt{5}}{2*1}=\frac{17-5\sqrt{5}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+5\sqrt{5}}{2*1}=\frac{17+5\sqrt{5}}{2}
$