51^2+140^2=c^2

Solution for 51^2+140^2=c^2 equation:

51^2+140^2=c^2

We move all terms to the left:

51^2+140^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+22201=0

a = -1; b = 0; c = +22201;
Δ = b2-4ac
Δ = 02-4·(-1)·22201
Δ = 88804
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{88804}=298$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-298}{2*-1}=\frac{-298}{-2}
=+149
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+298}{2*-1}=\frac{298}{-2}
=-149
$