20^2+21^2=c^2

Solution for 20^2+21^2=c^2 equation:

20^2+21^2=c^2

We move all terms to the left:

20^2+21^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+841=0

a = -1; b = 0; c = +841;
Δ = b2-4ac
Δ = 02-4·(-1)·841
Δ = 3364
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3364}=58$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-58}{2*-1}=\frac{-58}{-2}
=+29
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+58}{2*-1}=\frac{58}{-2}
=-29
$