x^2+32x-160=0

Solution for x^2+32x-160=0 equation:

x^2+32x-160=0

a = 1; b = 32; c = -160;
Δ = b2-4ac
Δ = 322-4·1·(-160)
Δ = 1664
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:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8\sqrt{26}}{2*1}=\frac{-32-8\sqrt{26}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8\sqrt{26}}{2*1}=\frac{-32+8\sqrt{26}}{2}
$