21^2+28^2=c^2

Solution for 21^2+28^2=c^2 equation:

21^2+28^2=c^2

We move all terms to the left:

21^2+28^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+1225=0

a = -1; b = 0; c = +1225;
Δ = b2-4ac
Δ = 02-4·(-1)·1225
Δ = 4900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4900}=70$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-70}{2*-1}=\frac{-70}{-2}
=+35
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+70}{2*-1}=\frac{70}{-2}
=-35
$