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Let $$f(x)= [x]^{2}-[x+3]-3, x\in \mathbb R$$, where $$[\cdot]$$ is the greatest integer funtion. Then
$$f(x)=[x]^2-[x+3]-3$$
$$f(x)=[x]^2-[x]-6$$
$$=([x]-3)([x]+2)$$
$$f(x)=0 \text{ for } x \in[-2,1) \cup[3,4)$$
$$f(x)>0$$:
$$\Rightarrow([x]-3)([x]+2)>0$$
$$\Rightarrow[x] \in(-\infty,-2) \cup(3, \infty)$$
$$\Rightarrow x \in(-\infty,-2) \cup[4, \infty)$$
$$f(x)<0$$:
$$\Rightarrow x \in[-1,3)$$
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