What is an easy and fast way to solve the problem without going through all these possibilities: a) $n^2-9n+20>0, 16-n^2>0$, b) $n^2-9n+20>0, 16-n^2<0$, c) $n^2-9n+20<0, 16-n^2>0$, and d) $n^2-9n+20<0, 16-n^2<0$? I got the correct answer, but it was very time consuming.
Find the product of all real numbers $/n^2-9n+20/=/16-n^2/$.