11^2+21^2=c^2

Solution for 11^2+21^2=c^2 equation:

11^2+21^2=c^2

We move all terms to the left:

11^2+21^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+562=0

a = -1; b = 0; c = +562;
Δ = b2-4ac
Δ = 02-4·(-1)·562
Δ = 2248
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:

$\sqrt{\Delta}=\sqrt{2248}=\sqrt{4*562}=\sqrt{4}*\sqrt{562}=2\sqrt{562}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{562}}{2*-1}=\frac{0-2\sqrt{562}}{-2}
=-\frac{2\sqrt{562}}{-2}
=-\frac{\sqrt{562}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{562}}{2*-1}=\frac{0+2\sqrt{562}}{-2}
=\frac{2\sqrt{562}}{-2}
=\frac{\sqrt{562}}{-1}
$