Given $x_0=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and $x_{n+1}=\frac{x_n^2-5}{2(x_n+2)}$, how can I find $x_n$? 
Let $x_0=\sqrt{2}+\sqrt{3}+\sqrt{6}$ and
$$\forall n \in \mathbb{N} : x_{n+1}=\dfrac{x_n^2-5}{2(x_n+2)}$$
Then $x_n=?$


My Try :
$$x_{n+1}(2x_n+4)=x_n^2-5 \\2x_{n+1}x_n+4x_{n+1}=x_n^2-5\\x_n^2-2x_{n+1}x_n-(4x_{n+1}+5)=0$$
So we have :
$$x_n=\dfrac{2x_{n+1}\pm\sqrt{2x_{n+1}^2+4(4x_{n+1}+5)}}{2}$$
Now what ?

A crucial observation is that the substitution $x_n=a_n-2$ brings the given recurrence relation into a familiar one, namely
$$ a_{n+1} = \frac{a_n^2-1}{2a_n} $$
which is associated with the duplication formula for the cotangent function.  In particular, $a_0=-\cot\theta$ implies $a_n=-\cot(2^n\theta)$ and $x_n=-2-\cot(2^n\theta)$. Since $a_0=\cot\frac{\pi}{24}$ we have
$$ x_n = -2+\cot\left(\frac{\pi\cdot2^n}{24}\right) $$
and our sequence keeps oscillating between $-2-\frac{1}{\sqrt{3}}$ and $-2+\frac{1}{\sqrt{3}}$ from $x_3$ on.

Anyway, for any $x_0\in\mathbb{R}$ the given sequence is never converging, since, as remarked in the comments, it is Newton's method applied to a quadratic polynomial with a negative discriminant ($x^2+4x+5$).
