w^2-4w-41=0

Solution for w^2-4w-41=0 equation:

w^2-4w-41=0

a = 1; b = -4; c = -41;
Δ = b2-4ac
Δ = -42-4·1·(-41)
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{5}}{2*1}=\frac{4-6\sqrt{5}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{5}}{2*1}=\frac{4+6\sqrt{5}}{2}
$