2x^2+32x-320=0

Solution for 2x^2+32x-320=0 equation:

2x^2+32x-320=0

a = 2; b = 32; c = -320;
Δ = b2-4ac
Δ = 322-4·2·(-320)
Δ = 3584
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:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{14}}{2*2}=\frac{-32-16\sqrt{14}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{14}}{2*2}=\frac{-32+16\sqrt{14}}{4}
$