64=a^2+a^2

Solution for 64=a^2+a^2 equation:

64=a^2+a^2

We move all terms to the left:

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

We get rid of parentheses

-a^2-a^2+64=0

We add all the numbers together, and all the variables

-2a^2+64=0

a = -2; b = 0; c = +64;
Δ = b2-4ac
Δ = 02-4·(-2)·64
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*-2}=\frac{0-16\sqrt{2}}{-4}
=-\frac{16\sqrt{2}}{-4}
=-\frac{4\sqrt{2}}{-1}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*-2}=\frac{0+16\sqrt{2}}{-4}
=\frac{16\sqrt{2}}{-4}
=\frac{4\sqrt{2}}{-1}
$