Inspired by Aareyan Manzoor

Nice question... :-} $\large f(x)=x^3+x^2-5x-1$ Since f(x) is continuous everywhere on $\mathcal{R}$ we can apply Intermediate value theorem on it to claim:
(1). f(-3)<0 and f(-2)>0 $\Rightarrow$ 1st root of f(x) lies between -3 and -2.
(2). f(-1)>0 and f(0)<0 $\Rightarrow$ 2nd root of f(x) lies between -1 and 0.
(3). f(1)<0 and f(2)>0 $\Rightarrow$ 3rd root of f(x) lies between 2 and 3.
When finding their $\lfloor~\rfloor$ , we get -3,-1 and 1 respectively the sum of which is $\Large \boxed{\color{#007fff}{-3}}$