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 $a\le -1 \vee a\ge 1$.