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.  

Here's a way
to get bounds on $x$
for a generalized version of this.
We want to solve
$x
=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor
$.
Upper bound:
If
$a = \max(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\le\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\le n \sqrt{x+a} \\
\text{so}\\
x^2
&\le n^2(x+a)\\
\end{array}
$
Therefore
$x^2-n^2x+n^4/4
\le n^2(a+n^2/4)
$
or
$|x-n^2/2|
\le n\sqrt{a+n^2/4}
= (n/2)\sqrt{4a+n^2}
$
so
$x
\le (n/2)(n+\sqrt{4a+n^2})
$.
For this case,
$n=3, a=2$,
so
$x
\le (3/2)(3+\sqrt{17})
\lt 10.7
$
so
$x \le 10$.
Lower bound:
If
$a = \min(a_k)$,
then
$\begin{array}\\
x
&=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\
&\ge\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\
&=\lfloor n \sqrt{x+a} \rfloor\\
&\gt n \sqrt{x+a}-1 \\
\text{so}\\
(x+1)^2
&\gt n^2(x+a)\\
&= n^2(x+1+a-1)\\
\end{array}
$
Therefore
$(x+1)^2-n^2(x+1)+n^4/4
\gt n^2(a-1+n^2/4)
$
or
$|x+1-n^2/2|
\gt n\sqrt{a-1+n^2/4}
= (n/2)\sqrt{4a-4+n^2}
$
so
$x+1-n^2/2
\gt (n/2)\sqrt{4a-4+n^2}
$
or
$x+1-n^2/2
\lt -(n/2)\sqrt{4a-4+n^2}
$.
Therefore
$x
\gt n^2/2-1+(n/2)\sqrt{4a-4+n^2}
$
or
$x
\lt n^2/2-1-(n/2)\sqrt{4a-4+n^2}
$.
For this case,
$n=3, a=0$,
so
$(n/2)\sqrt{4a-4+n^2}
=\frac32\sqrt{-4+9}
=\frac32\sqrt{5}
\approx 3.35
$
and
$n^2/2-1
=3.5
$
so
$x > 6.85$
or
$x < 0$.
Therefore
$x \ge 7$,
so that,
combining the bounds,
$7 \le x \le 10$.
