8x^2+32x-64=0.

Solution for 8x^2+32x-64=0. equation:

8x^2+32x-64=0.

a = 8; b = 32; c = -64;
Δ = b2-4ac
Δ = 322-4·8·(-64)
Δ = 3072
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:

$\sqrt{\Delta}=\sqrt{3072}=\sqrt{1024*3}=\sqrt{1024}*\sqrt{3}=32\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{3}}{2*8}=\frac{-32-32\sqrt{3}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{3}}{2*8}=\frac{-32+32\sqrt{3}}{16}
$