Evaluating $\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}$. I was wondering if it was possible to evaluate

$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}$$

I let the expression equal $x>0$ and wrote $$x=\sqrt{9-5\sqrt{3-x}}$$ However, there is not just one value $x$ can take; $x=2$ or $x=3$.
How do I find out which one it is, or does this infinite-nested radical converge at all? Perhaps it merely oscillates between $2$ and $3$, but I am not entirely sure. Any help or hints would be much appreciated.
Thank you in advance.

The ellipsis means "and so on". It measures the following: $$\sqrt{9-5}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5}}}$$ $$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5}}}}}$$ $$\vdots$$
Incidentally, I did not refuse to clarify the meaning. I am only active on Math.SE for so long. Whatever requests that occur can only be followed up the moment I am active, can see them and have time to act.

The answer is $2$. While $3$ is also a fixed point, it is unstable because if we let $x=3-\epsilon$ for some small $\epsilon$, and iterate $x\leftarrow \sqrt{9-5\sqrt{3-x}}$, it will diverge away from $3$.
$\sqrt{9-5\sqrt{3-x}}$">
If you look at the graph, you will find that the slope approaches $\infty$ as $x\to 3$. The derivative of $\sqrt{9-5\sqrt{3-x}}$ is $\frac5{4\sqrt{9-5\sqrt{3-x}}\sqrt{3-x}}$. When $x\to 3$, the $\sqrt{3-x}$ in the denominator will approach $0$, which means the derivative approaches $\infty$ as $x\to 3$. Therefore, the fixed point is unstable and will very quickly diverge away from $3$. Plugging $2$ into the equation gives $\frac58$, which is less than $1$. Therefore, the fixed point is stable. 
In conclusion:

$$\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-5\sqrt{3-\sqrt{9-\cdots}}}}}}}=2$$

