324+225=c^2

Solution for 324+225=c^2 equation:

324+225=c^2

We move all terms to the left:

324+225-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+549=0

a = -1; b = 0; c = +549;
Δ = b2-4ac
Δ = 02-4·(-1)·549
Δ = 2196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2196}=\sqrt{36*61}=\sqrt{36}*\sqrt{61}=6\sqrt{61}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{61}}{2*-1}=\frac{0-6\sqrt{61}}{-2}
=-\frac{6\sqrt{61}}{-2}
=-\frac{3\sqrt{61}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{61}}{2*-1}=\frac{0+6\sqrt{61}}{-2}
=\frac{6\sqrt{61}}{-2}
=\frac{3\sqrt{61}}{-1}
$