w^2+35w-2=0

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

w^2+35w-2=0

a = 1; b = 35; c = -2;
Δ = b2-4ac
Δ = 352-4·1·(-2)
Δ = 1233
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:

$\sqrt{\Delta}=\sqrt{1233}=\sqrt{9*137}=\sqrt{9}*\sqrt{137}=3\sqrt{137}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-3\sqrt{137}}{2*1}=\frac{-35-3\sqrt{137}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+3\sqrt{137}}{2*1}=\frac{-35+3\sqrt{137}}{2}
$