2x^2+20x-160=0

Solution for 2x^2+20x-160=0 equation:

2x^2+20x-160=0

a = 2; b = 20; c = -160;
Δ = b2-4ac
Δ = 202-4·2·(-160)
Δ = 1680
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:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{105}}{2*2}=\frac{-20-4\sqrt{105}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{105}}{2*2}=\frac{-20+4\sqrt{105}}{4}
$