b^2+4b-41=0

Solution for b^2+4b-41=0 equation:

b^2+4b-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:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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