2x^2+20x-2400=0

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

2x^2+20x-2400=0

a = 2; b = 20; c = -2400;
Δ = b2-4ac
Δ = 202-4·2·(-2400)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-140}{2*2}=\frac{-160}{4}
=-40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+140}{2*2}=\frac{120}{4}
=30
$