n^2+20n-2400=0

Solution for n^2+20n-2400=0 equation:

n^2+20n-2400=0

a = 1; b = 20; c = -2400;
Δ = b2-4ac
Δ = 202-4·1·(-2400)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-100}{2*1}=\frac{-120}{2}
=-60
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+100}{2*1}=\frac{80}{2}
=40
$