(2x^2)-(32x)+8=0

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

(2x^2)-(32x)+8=0

a = 2; b = -32; c = +8;
Δ = b2-4ac
Δ = -322-4·2·8
Δ = 960
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:

$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{15}}{2*2}=\frac{32-8\sqrt{15}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{15}}{2*2}=\frac{32+8\sqrt{15}}{4}
$