2x^2+16x-9920=0

Solution for 2x^2+16x-9920=0 equation:

2x^2+16x-9920=0

a = 2; b = 16; c = -9920;
Δ = b2-4ac
Δ = 162-4·2·(-9920)
Δ = 79616
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:

$\sqrt{\Delta}=\sqrt{79616}=\sqrt{256*311}=\sqrt{256}*\sqrt{311}=16\sqrt{311}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{311}}{2*2}=\frac{-16-16\sqrt{311}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{311}}{2*2}=\frac{-16+16\sqrt{311}}{4}
$