w^2+10w-120=0

Solution for w^2+10w-120=0 equation:

w^2+10w-120=0

a = 1; b = 10; c = -120;
Δ = b2-4ac
Δ = 102-4·1·(-120)
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{145}}{2*1}=\frac{-10-2\sqrt{145}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{145}}{2*1}=\frac{-10+2\sqrt{145}}{2}
$