2x^2+32x=64

Solution for 2x^2+32x=64 equation:

2x^2+32x=64

We move all terms to the left:

2x^2+32x-(64)=0

a = 2; b = 32; c = -64;
Δ = b2-4ac
Δ = 322-4·2·(-64)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{6}}{2*2}=\frac{-32-16\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{6}}{2*2}=\frac{-32+16\sqrt{6}}{4}
$