p^2+120p-11000=0

Solution for p^2+120p-11000=0 equation:

p^2+120p-11000=0

a = 1; b = 120; c = -11000;
Δ = b2-4ac
Δ = 1202-4·1·(-11000)
Δ = 58400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{58400}=\sqrt{400*146}=\sqrt{400}*\sqrt{146}=20\sqrt{146}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-20\sqrt{146}}{2*1}=\frac{-120-20\sqrt{146}}{2}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+20\sqrt{146}}{2*1}=\frac{-120+20\sqrt{146}}{2}
$