I am tasked to find the integral part of $\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}$, where the number of $2018$ is $2018$.  My attempt:  I came up with an upper bound, which was $\sqrt{2018+\sqrt{2018+\sqrt{2018+...}}}$, where there were infinite number of $2018$.  I let this be $a$, and then I have $a^2=2018+a$, giving me $45.4249374...$.  My problem here is that I have no idea what to set as a lower bound.  Can anyone give me a hint to solve this?  Alternatively, another approach could also be provided if it is more efficient.
Could you please help me computing    $\Big\lfloor\underbrace{\sqrt{2017+\sqrt{2017+...+\sqrt{2017}}}}_\text{2017 roots}\Big\rfloor$?
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