24^2+26^2=c^2

Solution for 24^2+26^2=c^2 equation:

24^2+26^2=c^2

We move all terms to the left:

24^2+26^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+1252=0

a = -1; b = 0; c = +1252;
Δ = b2-4ac
Δ = 02-4·(-1)·1252
Δ = 5008
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:

$\sqrt{\Delta}=\sqrt{5008}=\sqrt{16*313}=\sqrt{16}*\sqrt{313}=4\sqrt{313}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{313}}{2*-1}=\frac{0-4\sqrt{313}}{-2}
=-\frac{4\sqrt{313}}{-2}
=-\frac{2\sqrt{313}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{313}}{2*-1}=\frac{0+4\sqrt{313}}{-2}
=\frac{4\sqrt{313}}{-2}
=\frac{2\sqrt{313}}{-1}
$