32=4r^2

Solution for 32=4r^2 equation:

32=4r^2

We move all terms to the left:

32-(4r^2)=0

a = -4; b = 0; c = +32;
Δ = b2-4ac
Δ = 02-4·(-4)·32
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*-4}=\frac{0-16\sqrt{2}}{-8}
=-\frac{16\sqrt{2}}{-8}
=-\frac{2\sqrt{2}}{-1}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*-4}=\frac{0+16\sqrt{2}}{-8}
=\frac{16\sqrt{2}}{-8}
=\frac{2\sqrt{2}}{-1}
$