<s>[INST] Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
 = 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
 = 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
 [/INST] Well, you know that $$\frac{(n+4)!}{(n+2)!} = (n+4)\cdot(n+3) = n^2+7n+12=90$$$$n^2+7n-78=0$$The solutions to this quadratic are $n=6$ and $n=-13$, so we take $n=6$ to be our answer.
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