20^2+25^2=c^2

Solution for 20^2+25^2=c^2 equation:

20^2+25^2=c^2

We move all terms to the left:

20^2+25^2-(c^2)=0

We add all the numbers together, and all the variables

-1c^2+1025=0

a = -1; b = 0; c = +1025;
Δ = b2-4ac
Δ = 02-4·(-1)·1025
Δ = 4100
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:

$\sqrt{\Delta}=\sqrt{4100}=\sqrt{100*41}=\sqrt{100}*\sqrt{41}=10\sqrt{41}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{41}}{2*-1}=\frac{0-10\sqrt{41}}{-2}
=-\frac{10\sqrt{41}}{-2}
=-\frac{5\sqrt{41}}{-1}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{41}}{2*-1}=\frac{0+10\sqrt{41}}{-2}
=\frac{10\sqrt{41}}{-2}
=\frac{5\sqrt{41}}{-1}
$