<s>[INST] If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval 
$(A)(-1,0)\cup(0,1)$
$(B)(1,2)$ 
$(C)(-2,-1)$
$(D)(-\infty,-2)\cup(2,\infty)$

My try:
Let $x-\lfloor x \rfloor  = \{x\}= t$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq t<1$. Then
$$\Rightarrow -3t^2+2t+a^2 = 0\Rightarrow 3t^2-2t-a^2 = 0$$
$$\displaystyle \Rightarrow t = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
I do not know how to solve further.
 [/INST] EDIT (ELABORATION)
There is probably a typo here; it should say real solutions. 
Assume it has solutions. 
$$0\leq t<1,  t=\frac{1\pm \sqrt{1+3a^2}}{3} \Leftrightarrow 1 \le \sqrt{1+3a^2} < 2 $$
From the fact that $1-\sqrt{1+3a^2}<0$ if $a \neq 0$. From here, we square both sides to get $$0 \le a^2 \le 1$$
This implies $-1 \le a \le 1$. 
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