391^2+120^2=c^2

Solution for 391^2+120^2=c^2 equation:

391^2+120^2=c^2

We move all terms to the left:

391^2+120^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+167281=0

a = -1; b = 0; c = +167281;
Δ = b2-4ac
Δ = 02-4·(-1)·167281
Δ = 669124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{669124}=818$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-818}{2*-1}=\frac{-818}{-2}
=+409
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+818}{2*-1}=\frac{818}{-2}
=-409
$