x^2+32x-320=0

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

x^2+32x-320=0

a = 1; b = 32; c = -320;
Δ = b2-4ac
Δ = 322-4·1·(-320)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-48}{2*1}=\frac{-80}{2}
=-40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+48}{2*1}=\frac{16}{2}
=8
$