32-2x^2+8x=0

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

32-2x^2+8x=0

a = -2; b = 8; c = +32;
Δ = b2-4ac
Δ = 82-4·(-2)·32
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*-2}=\frac{-8-8\sqrt{5}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*-2}=\frac{-8+8\sqrt{5}}{-4}
$