a^2+(2a)^2=72^2

Solution for a^2+(2a)^2=72^2 equation:

a^2+(2a)^2=72^2

We move all terms to the left:

a^2+(2a)^2-(72^2)=0

We add all the numbers together, and all the variables

3a^2-5184=0

a = 3; b = 0; c = -5184;
Δ = b2-4ac
Δ = 02-4·3·(-5184)
Δ = 62208
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:

$\sqrt{\Delta}=\sqrt{62208}=\sqrt{20736*3}=\sqrt{20736}*\sqrt{3}=144\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-144\sqrt{3}}{2*3}=\frac{0-144\sqrt{3}}{6}
=-\frac{144\sqrt{3}}{6}
=-24\sqrt{3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+144\sqrt{3}}{2*3}=\frac{0+144\sqrt{3}}{6}
=\frac{144\sqrt{3}}{6}
=24\sqrt{3}
$