2x^2+32x+64=0

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

2x^2+32x+64=0

a = 2; b = 32; c = +64;
Δ = b2-4ac
Δ = 322-4·2·64
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{2}}{2*2}=\frac{-32-16\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{2}}{2*2}=\frac{-32+16\sqrt{2}}{4}
$