a^2+32^2=40^2

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

a^2+32^2=40^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

a^2-576=0

a = 1; b = 0; c = -576;
Δ = b2-4ac
Δ = 02-4·1·(-576)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*1}=\frac{-48}{2}
=-24
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*1}=\frac{48}{2}
=24
$