w^2+35w+100=0

Solution for w^2+35w+100=0 equation:

w^2+35w+100=0

a = 1; b = 35; c = +100;
Δ = b2-4ac
Δ = 352-4·1·100
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-5\sqrt{33}}{2*1}=\frac{-35-5\sqrt{33}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+5\sqrt{33}}{2*1}=\frac{-35+5\sqrt{33}}{2}
$