a^2+(9a)^2=65^2

Solution for a^2+(9a)^2=65^2 equation:

a^2+(9a)^2=65^2

We move all terms to the left:

a^2+(9a)^2-(65^2)=0

We add all the numbers together, and all the variables

10a^2-4225=0

a = 10; b = 0; c = -4225;
Δ = b2-4ac
Δ = 02-4·10·(-4225)
Δ = 169000
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:

$\sqrt{\Delta}=\sqrt{169000}=\sqrt{16900*10}=\sqrt{16900}*\sqrt{10}=130\sqrt{10}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-130\sqrt{10}}{2*10}=\frac{0-130\sqrt{10}}{20}
=-\frac{130\sqrt{10}}{20}
=-\frac{13\sqrt{10}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+130\sqrt{10}}{2*10}=\frac{0+130\sqrt{10}}{20}
=\frac{130\sqrt{10}}{20}
=\frac{13\sqrt{10}}{2}
$