2x2+800x-19200=0

Solution for 2x2+800x-19200=0 equation:

2x^2+800x-19200=0

a = 2; b = 800; c = -19200;
Δ = b2-4ac
Δ = 8002-4·2·(-19200)
Δ = 793600
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:

$\sqrt{\Delta}=\sqrt{793600}=\sqrt{25600*31}=\sqrt{25600}*\sqrt{31}=160\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(800)-160\sqrt{31}}{2*2}=\frac{-800-160\sqrt{31}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(800)+160\sqrt{31}}{2*2}=\frac{-800+160\sqrt{31}}{4}
$