Figuring domain of constant $a$ in a equation with some condition Here is what questions says

Question: If $a\in \mathrm{R}$ and the equation $-3(x-[x])^2+2(x-[x])+a^2=0$ (where $[\cdot]$ denotes the greatest integer $\leq x$) has no integral solutions, then all possible values of $a$ lie in which interval?

I have no idea how to get started with this one. I have seen problems like no real solution $(D\lt 0)$, both real solutions $(D\ge 0)$, equal solutions $(D=0)$ and all but no integral solutions is something new to me. What all I can do here is:
$$-3(x-[x])^2+2(x-[x])+a^2=0$$
$$\implies -3\{x\}^2+2\{x\}+a^2=0 $$
where $\{x\}$ is the fractional part function. Taking (just for convenience) $\{x\}=X$
$$\implies 3X^2-2X-a^2=0$$
$$\implies X^2-\frac{2}{3}X-\frac{a^2}{3}=0$$
That's all what I can do here. Any hints for further solving?
EDIT: In given question, there are some options also. I am adding them here.

$(a)\ (-\infty,\ -2)\cup (2,\ \infty)$
$(b)\ (-1,\ 0)\cup (0,\ 1)$
$(c)\ (1,\ 2)$
$(d)\ (-2,\ -1)$

And now please don't tell me that I can take values from every option (lying in the domain of the option) and put that in my equation and solve them individually. That won't be a good idea while doing problems for practice.
 A: If $a^2 $ is not integral then $x$ must not be integral.
So assume that $a^2$ is nonnegative integer. If $$x=n+\alpha,\ 0\leq
\alpha <1,\ n\in \mathbb{Z}$$ then $$ -3 \alpha^2+2\alpha +a^2=0 $$
So $$ \alpha = \frac{- 1\pm \sqrt{ 1+3a^2 }}{-3} \Rightarrow \alpha
= \frac{1+\sqrt{1+3a^2}}{3}\ {\text or}\ a=0,\ \alpha=0 $$
So $a^2< 1$. So $a=0,\ \alpha=2/3$ But as we have shown that if $a=0$ then there exists integral solution.
So if $a^2$ is not nonneagtive integer, the equation has no
integral solution. So (b).
A: One has $x-\lfloor x\rfloor=0$ iff $x$ is an integer. When the equation
$$3\bigl(x-\lfloor x\rfloor\bigr)^2-2\bigl(x-\lfloor x\rfloor\bigr)-a^2=0\tag{1}$$
has integer solutions $x$ then necessarily $a^2=0$, whence $a=0$. Conversely, if $a=0$ then $(1)$ can be rewritten as
$$\bigl(x-\lfloor x\rfloor\bigr)\bigl(3(x-\lfloor x\rfloor)-2\bigr)=0\ ,$$
which is obviously solved by arbitrary integer $x$.
It follows that $(1)$ has no integer solutions iff $a\ne0$, which can be written as $a\in{\mathbb R}\setminus\{0\}$. None of the given options coincides with this.
