Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$ 
I tried to solve this equation.
   First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program.  I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.  

You can use inequalities to simplify your problem. 
Since $\lfloor x \rfloor \le x$. Therefore we've 
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\
&\le 3\sqrt {x+2}\\
\end{align}
$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$
This gives us the range $x \in [-1,10] \tag1$.
Also $\lfloor x \rfloor \ge x-1$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\ge  \sqrt x+\sqrt {x+1}+\sqrt{x+2} \color{red}{-1}\\
&\ge 3\sqrt {x}-1\\
\end{align}
$$\implies x+1 \ge 3\sqrt x$$
$$\implies x^2-7x+1\ge 0 \, ; x \in \mathbb Z$$
This gives us $x \in (-\infty, 0]\cup [7,\infty)\tag2$ 
Taking intersection of $(1)$ and $(2)$, and taking care of domain, I.e. $x\ge 0$, we get
$$\color{blue}{x \in \{0,7,8,9,10\}}$$
Now you can check for $x=0,7,8,9,10$, which is quite easy now.
