16^2+20^2=c^2

Solution for 16^2+20^2=c^2 equation:

16^2+20^2=c^2

We move all terms to the left:

16^2+20^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+656=0

a = -1; b = 0; c = +656;
Δ = b2-4ac
Δ = 02-4·(-1)·656
Δ = 2624
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:

$\sqrt{\Delta}=\sqrt{2624}=\sqrt{64*41}=\sqrt{64}*\sqrt{41}=8\sqrt{41}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{41}}{2*-1}=\frac{0-8\sqrt{41}}{-2}
=-\frac{8\sqrt{41}}{-2}
=-\frac{4\sqrt{41}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{41}}{2*-1}=\frac{0+8\sqrt{41}}{-2}
=\frac{8\sqrt{41}}{-2}
=\frac{4\sqrt{41}}{-1}
$