2x^2+40x-2400=0

Solution for 2x^2+40x-2400=0 equation:

2x^2+40x-2400=0

a = 2; b = 40; c = -2400;
Δ = b2-4ac
Δ = 402-4·2·(-2400)
Δ = 20800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20800}=\sqrt{1600*13}=\sqrt{1600}*\sqrt{13}=40\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{13}}{2*2}=\frac{-40-40\sqrt{13}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{13}}{2*2}=\frac{-40+40\sqrt{13}}{4}
$