64=4x^2+32x

Solution for 64=4x^2+32x equation:

64=4x^2+32x

We move all terms to the left:

64-(4x^2+32x)=0

We get rid of parentheses

-4x^2-32x+64=0

a = -4; b = -32; c = +64;
Δ = b2-4ac
Δ = -322-4·(-4)·64
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32\sqrt{2}}{2*-4}=\frac{32-32\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32\sqrt{2}}{2*-4}=\frac{32+32\sqrt{2}}{-8}
$