r^2+r^2=324

Solution for r^2+r^2=324 equation:

r^2+r^2=324

We move all terms to the left:

r^2+r^2-(324)=0

We add all the numbers together, and all the variables

2r^2-324=0

a = 2; b = 0; c = -324;
Δ = b2-4ac
Δ = 02-4·2·(-324)
Δ = 2592
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:

$\sqrt{\Delta}=\sqrt{2592}=\sqrt{1296*2}=\sqrt{1296}*\sqrt{2}=36\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36\sqrt{2}}{2*2}=\frac{0-36\sqrt{2}}{4}
=-\frac{36\sqrt{2}}{4}
=-9\sqrt{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36\sqrt{2}}{2*2}=\frac{0+36\sqrt{2}}{4}
=\frac{36\sqrt{2}}{4}
=9\sqrt{2}
$