2x^2+160x-6300=0

Solution for 2x^2+160x-6300=0 equation:

2x^2+160x-6300=0

a = 2; b = 160; c = -6300;
Δ = b2-4ac
Δ = 1602-4·2·(-6300)
Δ = 76000
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:

$\sqrt{\Delta}=\sqrt{76000}=\sqrt{400*190}=\sqrt{400}*\sqrt{190}=20\sqrt{190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-20\sqrt{190}}{2*2}=\frac{-160-20\sqrt{190}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+20\sqrt{190}}{2*2}=\frac{-160+20\sqrt{190}}{4}
$