#1**+1 **

Let's split this into two inequalities and solve both to find what x is.

We have

\(64 \leq n^2\)

and

\(n^2 \leq 100+20n\)

Let's solve the first equation. Square rooting both sides, we get

\(n \geq 8\) or \(n \leq -8\)

Solving the second equation, we get that

\(n^2-20n-100 \leq 0\\ 10\left(1-\sqrt{2}\right) \leq n\leq10\left(1+\sqrt{2}\right)\)

Merging overlapping intervals, we have

\(8\le \:n\le \:10\left(1+\sqrt{2}\right)\)

This rounds to

\(8\le \:n\le \:24.14\)

So our answer is 17.

Thanks! :)

NotThatSmart Jun 19, 2024