Find the solution of $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$ Is anyone able to help me with the following equation concerned the floor function  $\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$
I don't know how to deal with the floor terms properly.

$$\lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2=0$$
\begin{matrix}
   x-1 &\lt &\lfloor x\rfloor &\lt &x+1\\
   x^2-1 &\lt &\lfloor x^2\rfloor &\lt &x^2+1\\
   \\
   3x-1 &\lt &\lfloor 3x\rfloor &\lt &3x+1\\
   -3x-1 &\lt &-\lfloor 3x\rfloor &\lt &-3x+1\\
   \\
   x^2-3x-2 &\lt &\lfloor x^2 \rfloor-\lfloor 3x\rfloor &\lt &x^2-3x+2\\
   x^2-3x-2 &\lt &-2 &\lt &x^2-3x+2\\
   -2 &\lt &-x^2+3x-2 &\lt &2\\
   -2 &\lt &x^2-3x+2 &\lt &2\\
   0 &\lt &x &\lt &3
\end{matrix}
The places where $\lfloor x^2\rfloor$ changes value are
$\{1, \sqrt 2, \sqrt 3, 2, \sqrt 5, \sqrt 6, \sqrt 7, \sqrt 8, 3 \}$
The places where $\lfloor 3x\rfloor$ changes value are
$\{\frac 13, \frac 23, 1, \frac 43, \frac 53, 2, \frac 73, \frac 83, 3 \}$
The sorted list of the union of these two sets is
$$\left\{\frac 13, \frac 23, 1, \frac 43, \sqrt 2, \frac 53,\sqrt 3 , 2,
  \sqrt 5, \frac 73, \sqrt 6, \sqrt 7, \frac 83, \sqrt 8, 3 \right\}$$
\begin{matrix}
   \text{breakpoint} & \lfloor x^2\rfloor &\lfloor 3x\rfloor & \lfloor{x^2}\rfloor−\lfloor{3x}\rfloor+2\\
   \frac 13 & 0 & 1 & 1  &\\
   \frac 23 & 0 & 2 & 0  & \checkmark\\
   1        & 1 & 3 & 0  & \checkmark\\
   \frac 43 & 1 & 4 & -1 &\\
   \sqrt 2  & 2 & 4 & 0  & \checkmark\\
   \frac 53 & 2 & 5 & -1 &\\
   \sqrt 3  & 3 & 5 & 0  & \checkmark\\
   2        & 4 & 6 & 0  & \checkmark\\
   \sqrt 5  & 5 & 6 & 1 &\\
   \frac 73 & 5 & 7 & 0 &\checkmark\\
   \sqrt 6  & 6 & 7 & 1 &\\
   \sqrt 7  & 7 & 7 & 2 &\\
   \frac 83 & 7 & 8 & 1 &\\
   \sqrt 8  & 8 & 8 & 2 &\\
   3        & 9 & 9 & 2 &\\
\end{matrix}
Hence the solution set is
$$ x \in 
   \left[ 2/3, 4/3 \right) \cup
   \left[ \sqrt 2, 5/3 \right) \cup
   \left[ \sqrt 3, \sqrt 5 \right) \cup
   \left[ 7/3, \sqrt 6 \right)
$$
