[Solution] $x=-2$ does not satisfy the given equation. Also, according to the problem, if $x$ is not an integer, then $\{x\}=[x]+1$. Thus, the given equation can be transformed into $\{x\}^{2}+4\{x\}+4=0$, which simplifies to $(\{x\}+2)^{2}=0$,
yielding $\{x\}=-2$, hence $\quad-3<x<-2$. Therefore, the answer is $(D)$.
$33 \cdot 37\{x\}$ represents the smallest integer not less than $x$, $[x]$ represents the largest integer not greater than $x$. For example, $\{\pi\}=4,[\pi]=3$, then the solution to the equation $\{x\}^{2}+4([x]+1)+4=0$ is
(A) $x=-2$.
(B) $x-3$.
(D) $-3<x<-2$.
(6th "Jinyun Cup" Junior High School Mathematics Invitational, 1989)