Example 8 The number of solutions to the equation $x^{2}-2[x]-3=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
From the given equation, we get $[x]=\frac{x^{2}-3}{2}$.
And $x-1<[x] \leqslant x$, so $x-1<\frac{x^{2}-3}{2} \leqslant x$.
Solving this, we get $-1 \leqslant x<1-\sqrt{2}$ or $1+\sqrt{2}<x \leqslant 3$.
Therefore, $[x]$ can only take the values $-1,2,3$.
When $[x]=-1$, $\frac{x^{2}-3}{2}=-1$, solving this gives $x=-1$;
When $[x]=2$, $\frac{x^{2}-3}{2}=2$, solving this gives $x=\sqrt{7}$;
When $[x]=3$, $\frac{x^{2}-3}{2}=3$, solving this gives $x=3$.
Therefore, the original equation has 3 solutions.
[Note] The solution methods for the above three examples are the same, and the solution steps can be summarized as:
(1) Solve $[x]$ from the original equation (expressed as an algebraic expression in terms of $x$), substitute it into the inequality system $x-1<[x] \leqslant x$, find the range of $x$, and thus determine the "possible values" of $[x]$.
(2) Substitute these "possible values" into the original equation to solve.
(3) Check. This is because in (1), substituting $[x]$ into the inequality system "expands" the range of $x$, and the roots must be verified.