a^2+a^2=40^2

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

a^2+a^2=40^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

2a^2-1600=0

a = 2; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·2·(-1600)
Δ = 12800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{2}}{2*2}=\frac{0-80\sqrt{2}}{4}
=-\frac{80\sqrt{2}}{4}
=-20\sqrt{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{2}}{2*2}=\frac{0+80\sqrt{2}}{4}
=\frac{80\sqrt{2}}{4}
=20\sqrt{2}
$