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.  

First solve the real inequalities
$$
x \le \sqrt{x}+\sqrt{x+1}+\sqrt{x+2} < x+1
\tag{1}$$
Solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}$ is numerically $9.8956$ and solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=x+1$ is numerically $7.9813$.  So solution of (1) is
$$
7.9813 < x \le 9.8956
$$
Finally, assume $x$ is an integer.  We get $x=8$ or $x=9$.
