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   

$(A)(-1,0)\cup(0,1)$  

$(B)(1,2)$  

$(C)(-2,-1)$  

$(D)(-\infty,-2)\cup(2,\infty)$  

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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.