19^2+31^2=c^2

Solution for 19^2+31^2=c^2 equation:

19^2+31^2=c^2

We move all terms to the left:

19^2+31^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+1322=0

a = -1; b = 0; c = +1322;
Δ = b2-4ac
Δ = 02-4·(-1)·1322
Δ = 5288
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:

$\sqrt{\Delta}=\sqrt{5288}=\sqrt{4*1322}=\sqrt{4}*\sqrt{1322}=2\sqrt{1322}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{1322}}{2*-1}=\frac{0-2\sqrt{1322}}{-2}
=-\frac{2\sqrt{1322}}{-2}
=-\frac{\sqrt{1322}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{1322}}{2*-1}=\frac{0+2\sqrt{1322}}{-2}
=\frac{2\sqrt{1322}}{-2}
=\frac{\sqrt{1322}}{-1}
$