Is there another way of solving this binomial problem? The problem is as follows:  in $$(1+x)^n$$ find $n$ such that the fifth coefficient is $70$ and the seventh coefficient equals $28$.
I put $$ \binom{n}{4}=70 $$ $$\binom{n}{6}=28 $$
I ended up with two polynomial equations:
$$ n^4-6n^3+11n^2-6n-1680=0  $$
$$ n^6-15n^5+85n^4-225n^3+274n^2-120n-20160=0 $$
and the solution to this binomial is $n=8$.
Is there another (simpler) way to solve this problem?

Divide $\binom{n}{4}$ by $\binom{n}{6}$, it will give us
$$
\frac{70}{28} = \frac{\binom{n}{4}}{\binom{n}{6}} = \frac{\frac{n!}{4!\left(n-4\right)!}}{\frac{n!}{6!\left(n-6\right)!}} = \frac{6!\left(n-6\right)!}{4!\left(n-4\right)!} = \frac{6\cdot5\cdot4!\left(n-6\right)!}{4!\left(n-4\right)\cdot\left(n-5\right)\cdot\left(n-6\right)!} = \frac{30}{\left(n-4\right)\left(n-5\right)}
$$
hence
$$
\left(n-4\right)\left(n-5\right) = 12 \Longrightarrow n^{2} -9n+8 = 0
$$
the solutions are $n=8$ and $n=1$, since we want a solution greater than $6$, we must have $n=8$.
