If $ -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$ has no integral solution, then $a$ is if $a$ is real number and $\displaystyle -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$.has no real integral solution,
Then all possible values of $a$ lie in the interval..
(where $\lfloor x \rfloor $ means floor function of $x$)
$\bf{My\; Solution::}$ Let $x-\lfloor x \rfloor  = \{x\}= y$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq y<1$. Then equation.. 
$$\Rightarrow -3y^2+2y+a^2 = 0\Rightarrow 3y^2-2y-a^2 = 0$$
$$\displaystyle \Rightarrow y = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
Now If equation has no real solution , Then $y<0 \;\; \cup \;\; y\geq 1$.
So $$\Rightarrow \frac{1\pm \sqrt{1+3a^2}}{3}<0\;\; \cup \;\; \frac{1\pm \sqrt{1+3a^2}}{3}\geq 1$$
Now how can I calculate value of $a,$ after that
Help me
Thanks

Since  $$\frac{1- \sqrt{1+3a^2}}{3}<0,\quad \forall a\neq 0$$
then you only need to find $a$ such that
$$\frac{1+\sqrt{1+3a^2}}{3}>1\iff \sqrt{1+3a^2}>2\iff a^2\ge 1\iff a\le -1 \vee a\ge 1.$$
So, your equation has no solutions for $a\le -1 \vee a\ge 1$.
