how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$

My answer is:

$$\begin{align}
&amp;A = 3 - \frac {2}{A}\\
\implies &amp;\frac {A^2-3A+2}{A}=0\\
\implies &amp;A^2-3A+2=0\\
\implies &amp;(A-1)\cdot(A-2)=0\\
\implies &amp;A=1\;\text{ or }\; A=2
\end{align}$$

I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
This continued fraction is the limit of 
the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.
Your two solutions are correct. You can verify them as follows.

$$\color{blue}{1} = 3 - 2=3-\frac{2}{\color{\red}{1}}\tag{1}$$
Now replace the red 1 with the blue 1, which equals to the right hand side in (1).

$$\color{blue}{2} = 3 - 1=3-\frac{\color{cyan}{2}}{\color{\red}{2}}\tag{2}$$
Now replace the red 2 with the blue 2, which equals to the right hand side in (2).

I heard from the internet that this is one of the ways Ramanujan created some of his equalities.

Now 2 questions for you, 
(1) what if you replace the cyan 2 with blue 2 and so on so forth?
(2) what if you replace the cyan 2 with blue 2 and then replace the red 2 with blue 2 and so on so forth?