4x^2+32x-256=0

Solution for 4x^2+32x-256=0 equation:

4x^2+32x-256=0

a = 4; b = 32; c = -256;
Δ = b2-4ac
Δ = 322-4·4·(-256)
Δ = 5120
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:

$\sqrt{\Delta}=\sqrt{5120}=\sqrt{1024*5}=\sqrt{1024}*\sqrt{5}=32\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{5}}{2*4}=\frac{-32-32\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{5}}{2*4}=\frac{-32+32\sqrt{5}}{8}
$