The given equation is $36 x^{2}-12 a x+\left(a^{2}-b^{2}\right)=0$

Comparing it with $A x^{2}+B x+C=0$, we get
$A=36, B=-12 a$ and $C=a^{2}-b^{2}$

$\therefore$ Discriminant,

$D=B^{2}-4 A C=(-12 a)^{2}-4 \times 36 \times\left(a^{2}-b^{2}\right)=144 a^{2}-144 a^{2}+144 b^{2}=144 b^{2}>0$

So, the given equation has real roots

Now, $\sqrt{D}=\sqrt{144 b^{2}}=12 b$

$\begin{array}{l}
\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-(-12 a)+12 b}{2 \times 36}=\frac{12(a+b)}{72}=\frac{a+b}{0} \\
\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-(-12 a)-12 b}{2 \times 36}=\frac{12(a-b)}{72}=\frac{a-b}{6}
\end{array}$

Hence, $\frac{a+b}{6}$ and $\frac{a-b}{6}$ are the roots of the given equation.