Thinking about a good name

First note that if $n$ satisfies the inequality, then so does $-n$ . For the expression to be negative, one of $(n^2-2)$ and $(n^2-20)$ must be negative. Equating each to zero to find critical points, we can see that $\sqrt{2}<2\leq n \leq 4 < \sqrt{20}$ gives us some solutions. So $n$ can be $\pm 2, \pm 3, \pm 4$ , so $6$ solutions.

Using the identity, ${ a }^{ 2 }-{ b }^{ 2 }=(a-b)(a+b)$ .

We can split the question into four terms.

$(x-\sqrt { 2 } )(x+\sqrt { 2 } )(x-\sqrt { 20 } )(x+\sqrt { 20 } )<0$ .

Using wavy- curve method of intervals,(on the number line),

We get the solutions to be in the interval, $(-\sqrt { 20 } ,-\sqrt { 2 } )\cup (\sqrt { 2 } ,\sqrt { 20 } )$ .

As, $\sqrt { 20 } \approx 4.472$ .

We get the solution set as $\{ 2,3,4,-2,-3,-4\}$ .

So, the number of solutions is 6.