2x^2+20x-6300=0

Solution for 2x^2+20x-6300=0 equation:

2x^2+20x-6300=0

a = 2; b = 20; c = -6300;
Δ = b2-4ac
Δ = 202-4·2·(-6300)
Δ = 50800
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:

$\sqrt{\Delta}=\sqrt{50800}=\sqrt{400*127}=\sqrt{400}*\sqrt{127}=20\sqrt{127}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{127}}{2*2}=\frac{-20-20\sqrt{127}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{127}}{2*2}=\frac{-20+20\sqrt{127}}{4}
$