w^2+35w-150=0

Solution for w^2+35w-150=0 equation:

w^2+35w-150=0

a = 1; b = 35; c = -150;
Δ = b2-4ac
Δ = 352-4·1·(-150)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{73}}{2*1}=\frac{-35-5\sqrt{73}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{73}}{2*1}=\frac{-35+5\sqrt{73}}{2}
$