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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

3a^2-64=0

a = 3; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·3·(-64)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{3}}{2*3}=\frac{0-16\sqrt{3}}{6}
=-\frac{16\sqrt{3}}{6}
=-\frac{8\sqrt{3}}{3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{3}}{2*3}=\frac{0+16\sqrt{3}}{6}
=\frac{16\sqrt{3}}{6}
=\frac{8\sqrt{3}}{3}
$