Question: The question is

> 
> Solve for $n$ in the equation:
>  $$\binom{n+2}{4}=6\binom{n}{2}$$
> 
> 
>

Answer: 
If $n\ge 2$ (to be sure that $n+2\ge 4$ and $n\ge 2$ in the given binomial coefficients, otherwise see the comment below), you can write the LHS as $$\dbinom{n+2}{4}=\frac{(n+2)!}{4!(n+2-4)!}=\frac{(n+2)(n+1)n(n-1)(n-2)!}{4!(n-2)!}=\frac{(n+2)(n+1)n(n-1)}{24}$$ and the RHS as $$6\dbinom{n}{2}=6\frac{n!}{2(n-2)!}=6\frac{n(n-1)(n-2)!}{2(n-2)!}=3n(n-1)$$ Equating these two you get $$\frac{(n+2)(n+1)n(n-1)}{24}=3n(n-1) \implies (n+2)(n+1)=72$$ which gives $n=7$ (or $n=-10$ but this should be rejected).
**hint**: $\binom{n}{2}=\dfrac{n(n-1)}{2}, \binom{4}{2} = 6, \binom{n+2}{4} = \dfrac{(n+2)!}{4!\cdot (n-2)!}= \dfrac{(n-1)n(n+1)(n+2)}{24}$