x^2+32x-256=0

Solution for x^2+32x-256=0 equation:

x^2+32x-256=0

a = 1; b = 32; c = -256;
Δ = b2-4ac
Δ = 322-4·1·(-256)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{2}}{2*1}=\frac{-32-32\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{2}}{2*1}=\frac{-32+32\sqrt{2}}{2}
$