2. Let $x \in R$ satisfy $x^{2}-4[x]+3=0$, then $x=$
2. From $x \leqslant[x]<x+1$,
$$
\Rightarrow x^{2}+3<4 x+4 \text {, and } x^{2}+3 \geqslant 4 x \text {, }
$$

Solving gives $2-\sqrt{5}<x<2+\sqrt{5}$.
Also, $x^{2}=4[x]-3 \in \mathbb{Z}$
$$
x^{2}<(2+\sqrt{5})^{2}<18 \text {. }
$$

So $x^{2} \leqslant 17 \Rightarrow-1 \leqslant[x] \leqslant 4$.
Substituting $[x]$ into the equation, we get $x=\sqrt{13}, 3, \sqrt{5}, 1$. After verification, $x=3, \sqrt{5}$ or 1.