Big Theta Proof on polynomial function
Your solution is fine.  There are multiple ways to prove this fact, all of them valid/correct.  So your approach is valid, and it's possible that the solution from your class is also valid: that's not a contradiction.

Let me give you one piece of advice for structuring your proof in the future, though.  In the argument you sketched in the question, you are "working backwards": you start from the statement you wish were true (that $c_1 n^2 \le 2 n^2 -4n + 7 \le c_2 n^2$), and then you try to derive what needs to be true about $c_1,c_2$ for your wish to hold true.  Experience has proven that that kind of reasoning is error-prone, though: it's easy for you to make a mistake along the way.  It is also hard for others to check.

When you write up your proof, I suggest you do the reasoning in the opposite direction.  Start from what you know is true, and then derive the implications of that, ending with the conclusion that $c_1 n^2 \le 2 n^2 -4n + 7 \le c_2 n^2$.  This better matches how people think and makes it easier for the reader to understand what's going on.  And the purpose of a proof is to communicate an idea to the reader, so the proof should be structured to make the reader's life easier: to make it as easy as possible for the reader to verify that the proof's reasoning is correct and valid.  It's just like a good book; good writing is chosen to make life good for the reader.  Proofs are the same way.  And, ultimately, this will benefit you too: in my experience, if you follow this approach, then you're less likely to make a subtle mistake in your proof.

So, a better proof outline would be to show that $n^2 \le 2n^2 - 4n + 7$ holds for all $n\ge 2$ (because such-and-such), and show that $2n^2 - 4n + 7 \le 2n^2$ holds for all $n\ge 2$ (because blah-de-blah), and then conclude that $2n^2 - 4n + 7 \in \Theta(n^2)$.  I realize this kind of proof is harder to write, because you need to know the value of $c_1,c_2$ to use before you can start writing down this style of proof.  But you know what?  That's OK.  You can do a side calculation, on scratch paper, to figure out what value of $c_1,c_2$ to use.  Don't show us that side calculation; just figure out what value of $c_1,c_2$ would be good ones to do, and then start the proof with you pulling $c_1=1,c_2=2$ magically out of thin air, and demonstrate that your choice is valid.  This style of proof will be better for you, and better for the reader.