Solution for continued fraction To solve the finite continued fraction $[2,....,2]$, can I use this solution. 
Let $y$ be the value of $[2,....,2]$.
Then $y=[2;y]=2+\frac{1}{y}$. 
So $y-2=\frac{1}{y}$.
Thus $y^2-2y-1=0$ 
My question is is my method correct. Is this approach suitable to solve $[2,....,2]$ 


Is this approach suitable to solve $[2,....,2]$ 

No, because your continued fraction is finite. Moreover, it is rational and the solutions of $y^2-2y-1=0$ are irrational. 
Notation. 
\begin{equation*}
x_{n}=a_0+\cfrac{1}{a_1+\cfrac{1}{\begin{array}{ccc}a_{2}+ & & \\& \ddots & \\& & +\cfrac{1}{a_{n-1}+\cfrac{1}{a_{n}}}\end{array}}}=
\left[ a_{0};a_{1},a_{2},\ldots ,a_{n}\right] =[2;\underset{n\text{
elements}}{\underbrace{2,\ldots ,2}}]=\frac{p_{n}}{q_{n}}.
\end{equation*}
We then have $x_{0}=[a_{0}]=\frac{p_{0}}{q_{0}}=\frac{a_{0}}{1}=\frac{2}{1}$
 with $(p_{0},q_{0})=(2,1)$; and $x_{1}=\left[ a_{0};a_{1}\right] =\frac{p_{1}}{q_{1}
}=a_{0}+\frac{1}{a_{1}}=\frac{a_{0}a_{1}+1}{a_{1}}=\frac{5}{2}$; with $
(p_{1},q_{1})=(5,2)$. 
For $n\geq 2$ the integers $p_{n},q_{n}$ satisfy the
fundamental recurrent relation
\begin{eqnarray*}
p_{n} &=&a_{n}p_{n-1}+p_{n-2}=2p_{n-1}+p_{n-2}, \\
q_{n} &=&a_{n}q_{n-1}+q_{n-2}=2q_{n-1}+q_{n-2}.
\end{eqnarray*}
The next two terms of the finite sequence $(x_n)$ are
\begin{eqnarray*}
x_{2} &=&\left[ a_{0};a_{1},a_{2}\right] =a_{0}+\frac{1}{a_{1}+1/a_{2}}=
\frac{p_{2}}{q_{2}}=\frac{2p_{1}+p_{0}}{2q_{1}+q_{0}}=\frac{2(5)+2}{2(2)+1}=
\frac{12}{5}, \\
x_{3} &=&\left[ a_{0};a_{1},a_{2},a_{3}\right] =a_{0}+\frac{1}{
a_{1}+1/(a_{2}+1/a_{3})}=\frac{p_{3}}{q_{3}}=\frac{2p_{2}+p_{1}}{2q_{2}+q_{1}
}=\frac{2(12)+5}{2(5)+2}=\frac{29}{12}, \\
&&\cdots 
\end{eqnarray*}
and
\begin{equation*}
(p_{2},q_{2})=(12,5),(p_{3},q_{3})=(29,12),\ldots .
\end{equation*}
