x^2+20x=2400

Solution for x^2+20x=2400 equation:

x^2+20x=2400

We move all terms to the left:

x^2+20x-(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:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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