express $a$ in terms of $b$ and $c$ 
Given that $$c=\frac{\sqrt{a+3b}}{a-3b}$$ express $a$ in terms of $b$ and $c$

My attempt,
\begin{align}c^2(a^2-6ab+9b^2)&=a+3b\\
c^2a^2+(-6bc^2-1)a+9b^2c^2-3b&=0\\
a&=\frac{-(-6bc^2-1)\pm \sqrt{(-6bc^2-1)^2-4c^2(9b^2c^2-3b)}}{2c^2}\\
a&=\frac{6bc^2+1\pm \sqrt{24bc^2+1}}{2c^2}\end{align}
My question: Is my answer correct?

A slightly different take on Yves Daoust's approach:
Let $\sqrt{a+3b}=u\ge0$, so that $a=u^2-3b$.  Then 
$$c={u\over u^2-6b}\implies cu^2-u-6bc=0$$
The quadratic formula gives
$$u={1\pm\sqrt{1+24bc^2}\over2c}$$
as the formal solutions.  It is convenient to rewrite them as
$$u={1+\sqrt{1+24bc^3}\over2c}\qquad\text{and}\qquad u={-12bc\over1+\sqrt{1+24bc^2}}$$
From this it's easy to see that $u$ has no (real) solutions if $24bc^2\lt-1$, while
$$u=
\begin{cases}
0&\quad\text{if}\quad c=0\\\\
\displaystyle{1+\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad b,c\gt0\\\\
\displaystyle{1-\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad c\lt0\le b\\\\
\displaystyle{1\pm\sqrt{1+24bc^3}\over2c}&\quad\text{if}\quad\displaystyle{-1\over24c^2}\le b\lt0\lt c
\end{cases}$$
Note that the final case describes two solutions except when $b={-1\over24c^2}$.
