w^2+90w-340=0

Solution for w^2+90w-340=0 equation:

w^2+90w-340=0

a = 1; b = 90; c = -340;
Δ = b2-4ac
Δ = 902-4·1·(-340)
Δ = 9460
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:

$\sqrt{\Delta}=\sqrt{9460}=\sqrt{4*2365}=\sqrt{4}*\sqrt{2365}=2\sqrt{2365}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-2\sqrt{2365}}{2*1}=\frac{-90-2\sqrt{2365}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+2\sqrt{2365}}{2*1}=\frac{-90+2\sqrt{2365}}{2}
$