<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] When you take the $90$ to the other side you need to subtract it. You'll get
$n^2 +7n +12 -90 = 0,$
$n^2 +7n -78 = 0,$
$(n - 6)(n+13) = 0,$
So $n = 6$ or $n = -13$.
Clearly $n = -13 $  doesn't make sense, so we conclude:
$n = 6$.
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