I was given this problem in the text book:
$$\frac{(n+4)!}{(n+2)!} = 6$$
$$n \in I $$
Since the textbook doesn't have the solution, I'm wondering if I'm right:
$$\frac{(n+4)!}{(n+2)!} \Rightarrow \frac{(n+4)(n+3)(n+2)!}{(n+2)!} \Rightarrow (n+4)(n+3)$$
$$\Rightarrow (n+4)(n+3)$$ $$\Rightarrow n^2+4n+3n+12$$ $$\Rightarrow n^2+7n+12$$
$$\Rightarrow n^2 + 7n + 12 = 6$$ $$\Rightarrow n^2+7n+6=0$$
$$\Rightarrow n^2+7n+6=0$$ $$\Rightarrow n^2 +6n + n + 6 = 0 $$
$$\Rightarrow n(n + 6)1(n+6) = 0$$
$$\Rightarrow (n+1)(n+6) = 0 $$
Then $n$ can be two possible values:
$$n+1=0 $$ $$n+6 = 0$$
$$n=-1 \ or \ n=-6$$
If I go back to the original equation and replace $n$ with $-1$:
$$\frac{(-1+4)!}{(-1+2)!} = \frac{(3)!}{(1)!} = \frac{3*2*1}{1} = \frac{6}{1}$$
If replace $n$ with $-6$:
$$\frac{(-6+4)!}{(-6+2)!} = \frac{(-2)!}{(-4)!} = \varnothing$$
So is $n = -1$ or am I wrong? There's no solution shown in the text book.
You've already checked that when $n=-1$, then you get the equality you want, so that is definitely a solution. You can ignore the case $n=-6$ because the factorial is not defined for $n<0$. Your solution is correct.
The reason why the other "solution" $n=-6$ showed up is because it is a solution to the equation $(n+4)(n+3)=6$, which allows $n$ to be any real number. However, in the original equation, simply writing down $(n+2)!$ already implicitly implies that $n \ge -2$, restricting the space of your possible solutions.
As noted in the comment by Winther, it is assumed that $n$ is an integer, so solving $(n+4)(n+3)=6$ simply requires finding two integers whose product is $6$ and whose difference is $1$. This is a little bit quicker.