r^2-8r-41=0

Solution for r^2-8r-41=0 equation:

r^2-8r-41=0

a = 1; b = -8; c = -41;
Δ = b2-4ac
Δ = -82-4·1·(-41)
Δ = 228
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{57}}{2*1}=\frac{8-2\sqrt{57}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{57}}{2*1}=\frac{8+2\sqrt{57}}{2}
$