$ (n-7)^2$ is $\Theta(n^2) $ Prove if it's true $$ (n-7)^2 \, \text{is} \, \Theta(n^2) $$
Is this correct?
So far I have:
$ (n-7)^2 \, \text{is} \, O(n^2) \\ 
 n^2 -14n +49 \, \text{is} \, O(n^2) \\ 
 \begin{align} n^2 -14n +49 & \le \, C \cdot n^2 \, , \, n \ge 1 \\ 
 & \le 50n^2 \end{align} \\ $
$ (n-7)^2 \, \text{is} \, \Omega(n^2) \\ 
 n^2 -14n +49 \, \text{is} \, \Omega(n^2) \\
\begin{align} n^2 -14n +49 \, & \ge \, C \cdot (n^2) \; ,n \ge 1 \\
& \ge -14 n^2 \end{align}$
This proves it true right? Must both proofs have the same constant? I'm very new with Asymptotics.

Easier: $(n-7)^2 > (\frac{n}{2})^2=\Omega(n^2), \ (n-7)^2 < (2n)^2=O(n^2)$ for $n$ large enough, hence $(n-7)^2 = \Theta(n^2)$
