w^2+4w-120=0

Solution for w^2+4w-120=0 equation:

w^2+4w-120=0

a = 1; b = 4; c = -120;
Δ = b2-4ac
Δ = 42-4·1·(-120)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*1}=\frac{-4-4\sqrt{31}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*1}=\frac{-4+4\sqrt{31}}{2}
$