32^2+40^2=x^2

Solution for 32^2+40^2=x^2 equation:

32^2+40^2=x^2

We move all terms to the left:

32^2+40^2-(x^2)=0

We add all the numbers together, and all the variables

-1x^2+2624=0

a = -1; b = 0; c = +2624;
Δ = b2-4ac
Δ = 02-4·(-1)·2624
Δ = 10496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{10496}=\sqrt{256*41}=\sqrt{256}*\sqrt{41}=16\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{41}}{2*-1}=\frac{0-16\sqrt{41}}{-2}
=-\frac{16\sqrt{41}}{-2}
=-\frac{8\sqrt{41}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{41}}{2*-1}=\frac{0+16\sqrt{41}}{-2}
=\frac{16\sqrt{41}}{-2}
=\frac{8\sqrt{41}}{-1}
$