For solving $$[x^2]+2[x]=3x,....(1)$$ where [.] denotes GIF/Floor function, we use the estimation rule that $F-1<[F] \le F$. We write $x^2-1 < [x^2] \le x^2$ and $2(x-1) < 2[x] \le 2x$. Adding them we get $$ x^2+2x-3 < 3x \le x^2+2x \implies x^2-x-3<0 ~\&~ x(x-1) \ge 0.$$ These two inequations yield $x\in \left(\frac{1-\sqrt{13}}{2}, 0\right]\cup \left[1, \frac{1+\sqrt{13}}{2}\right)\implies x \in (-1.3027,0] \cup [1,2.3027).$ From (1), we have $3x \in I \implies (i): x \in I, (ii): x \notin I.$
Case (i): the possibilities are $x=-1,0,1,2$, wherein only $x=0,1$ satisfy (1).
Case (ii): The posibilities are $x=-1/3,-2/3, 4/3, 5/3$, but out of these only $-2/3$ satisfies Eq. (1).
Finally, Eq.(1) has three roots $-2/3,0,1$. However, if plot $f(x)=[x^2]+2[x]-3x$, there appears to be a root in the left vicinity of $-1$, see below
The question is what is this root numerically? What could be other methods of finding all roots of Eq. (1).
Your plot is missing some discrete points. For example, $f(-1)=2$ and this is not represented in your graph. What you see at $-1$ is just $\lim_{x\to-1^{-}}f(x)=0$. Because for a small $\varepsilon>0$, $f(-1-\varepsilon)=1+2(-2)-3(-1-\varepsilon)=3\varepsilon$.
For definitively identifying all roots of $f$, note that $3x$ must be an integer. So any root has to be of the form $\frac{m}{3}$ where $m$ is an integer. There are only finitely many such numbers to check in, say $[-1.3027,2.3027]$.