Showing $7n^2 + n = \Omega(n)$

Question

I have the intution of how this is true but I am facing problem to write it mathematically. Can someone help me with this? $$7n^{2} + n=\Omega(n)$$

Answer 1

As $n^2 \geq 0$, we have that $7n^2 \geq 0$. Thus $n + 7n^2 \geq n$ for all $n$, and thus $n + 7n^2 = \Omega(n)$.

If you like to show constants $c$ and $N$ such that $(n + 7n^2) \geq cn$ for all $n > N$, then this shows that you can take $c = N = 1$.