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.  

This is a little plodding, but $x\in\mathbb{Z}$ and $x\ge0$ (required in order for $\sqrt x$ to be real) tells us
$$x\ge\lfloor\sqrt0+\sqrt1+\sqrt2\rfloor=\lfloor2.414\rfloor=2$$
which tells us
$$x\ge\lfloor\sqrt2+\sqrt3+\sqrt4\rfloor=\lfloor5.145\rfloor=5$$
which tells us
$$x\ge\lfloor\sqrt5+\sqrt6+\sqrt7\rfloor=\lfloor7.331\rfloor=7$$
which tells us
$$x\ge\lfloor\sqrt7+\sqrt8+\sqrt9\rfloor=\lfloor8.474\rfloor=8$$
Finally, we see that if $x\ge10$, then the function $f(x)=\sqrt x+\sqrt{x+1}+\sqrt{x+2}-x$ is negative, since $f(10)=-0.056997...$ and
$$f'(x)={1\over2\sqrt x}+{1\over2\sqrt{x+1}}+{1\over2\sqrt{x+2}}-1\lt{3\over2\sqrt 9}-1=-{1\over2}\lt0$$
so we cannot have $x\le\sqrt x+\sqrt{x+1}+\sqrt{x+2}$, which is required in order to have $x=\lfloor\sqrt x+\sqrt{x+1}+\sqrt{x+2}\rfloor$, for $x\ge10$.  This leaves the two possibilities $x=8$ and $9$, which do solve the equation.
