Let    $x-\left[x\right]=t$

$\therefore -3t^2+2t+a^2=0$

As $x$ is not an integer, $t\ne 0$     $\therefore a\ne 0$

For a root of $x$ to exist at least one of the roots of $t$ should be between $0$ and $1$.

Roots    $\mathrm{=}\frac{-2\pm \sqrt{4-\left(4\right)\left(-3\right)a^2}}{-6}$

              $=\frac{-2\pm \sqrt{4+12a^2}}{-6}$

              $=\frac{-1\pm \sqrt{1+3a^2}}{-3}=\frac{1\mp \sqrt{1+3a^2}}{3}$

As we observe, one root is surely less than zero,

i.e.,   $\frac{1-\sqrt{1+3a^2}}{3}$

$\therefore$   For a solution to exist,

       $\frac{1+\sqrt{1+3a^2}}{3}<1$

$\therefore 1+\sqrt{1+3a^2}<3$

$\therefore \sqrt{1+3a^2}<2$

$\therefore 1+3a^2<4$

$\therefore 3a^2<3$

$\therefore a^2<1$

$\therefore a\in \left(-1,0\right)\cup \left(0,1\right)$.     $\left[\because a\ne 0\right]$