To evaluate $\lfloor\sqrt{80}\rfloor$, I need to find the largest integer that is less than or equal to $\sqrt{80}$. I know that $\sqrt{81}=9$, so $\sqrt{80}$ must be slightly less than 9. Therefore, the largest integer that is less than or equal to $\sqrt{80}$ is 8. So $\lfloor\sqrt{80}\rfloor=8$. 
Evaluate $\lfloor\sqrt{80}\rfloor$.