I have done the proof until the point when $T(n) \leq cn^{\log7}$.

But when it comes to finding the value of constant $c$, I am getting stuck.

The given recurrence relation is $T(n) = 7T(n/2) + n^2$. 

Since we already calculated the solution above which is $cn^{\log 7}$.

Inductive step:

Now we have to prove that $T(n) \leq c n^{\log7}$ where $c$ is a positive constant.
If we consider that the solution holds good for $n/2$ then we can prove that it works  for $n$ also: 
$$T(n/2) \leq c(n/2)^{\log7}.$$
Substituting these values in the recurrence relation:

$$
\begin{align*}
T(n) &\leq 7c/(2)^{\log7} \times (n)^{\log7} + n^2 \\
     &\leq cn^{\log7}, \text{ since $7/(2)^{\log7}$  is constant so can be ignored and $cn^{\log7} \gg n^2$ for large $n$} \\
     &\leq cn^{\log7} \text{ assuming $c$ is a constant $\geq 1$.}
\end{align*}
$$

Finally to find constant $c$, 

$$(7/(2)^{\log7}) \times cn^{\log7} + n^2 \leq cn^{\log7}. $$

I am not able to find appropriate $c$ for which the condition holds true.