w^2+6w-41=0

Solution for w^2+6w-41=0 equation:

w^2+6w-41=0

a = 1; b = 6; c = -41;
Δ = b2-4ac
Δ = 62-4·1·(-41)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10\sqrt{2}}{2*1}=\frac{-6-10\sqrt{2}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10\sqrt{2}}{2*1}=\frac{-6+10\sqrt{2}}{2}
$