We can choose $c=12$ and $n_0=1$ for the definition of Big-Oh to hold.

There is not a magic formula to find the constants. However, you can usually guess the values or deduct them by simple algebraic modification of $T(n)$.

$$T(n) = 7n^2 + 5 \le 7n^2 + 5n^2 \le 12n^2$$

Please note the choice of $n_0$ and $c$ are not unique. Moreover, these constants are not necessarily integers. According to the definition, these constants are non-zero and positive real values. Of course, if $n$ represents the program input size, then it is imperative to be an integer.

Please also note I have deliberately asked you to "show" $T(n) \in \Omicron(n^2)$ and not "to prove." However, a proof generally would require finding a pair of $c$ and $n_0$ for the definition to hold. Of course, you should present your argument more formally. For example, the following would make reasonable proof. Notice I deliberately chose a different pair of $c$ and $n_0$ from those selected above.

Proof: Let's set $c = 10$. Now suppose that $7n^2 + 5 > 10n^2$. Then by algebraic simplification, we get $3n^2 < 5$, and thus $-\sqrt{\frac{5}{3}} < n < \sqrt{\frac{5}{3}}$. So by the counterpositive, $7n^2 + 5 \le 10n^2$ for all $n \ge \sqrt{\frac{5}{3}}$. So if we set $n_0 = \left \lceil \sqrt{\frac{5}{3}} \right \rceil = 2$, we have proved that $7n^2 + 5 \le cn^2$ for $c = 10$ for all $n > n_0$. Thus we have proved that $T(n) \in \Omicron(n^2)$. Q.E.D.