2x^2+20x-8400=0

Solution for 2x^2+20x-8400=0 equation:

2x^2+20x-8400=0

a = 2; b = 20; c = -8400;
Δ = b2-4ac
Δ = 202-4·2·(-8400)
Δ = 67600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{67600}=260$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-260}{2*2}=\frac{-280}{4}
=-70
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+260}{2*2}=\frac{240}{4}
=60
$