r^2-32^2+12=0

Solution for r^2-32^2+12=0 equation:

r^2-32^2+12=0

We add all the numbers together, and all the variables

r^2-1012=0

a = 1; b = 0; c = -1012;
Δ = b2-4ac
Δ = 02-4·1·(-1012)
Δ = 4048
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:

$\sqrt{\Delta}=\sqrt{4048}=\sqrt{16*253}=\sqrt{16}*\sqrt{253}=4\sqrt{253}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{253}}{2*1}=\frac{0-4\sqrt{253}}{2}
=-\frac{4\sqrt{253}}{2}
=-2\sqrt{253}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{253}}{2*1}=\frac{0+4\sqrt{253}}{2}
=\frac{4\sqrt{253}}{2}
=2\sqrt{253}
$