$T(n)=7T\left ( \frac{n}{3} \right )\dotplus n^{2}$
Given recurrence relation is in this below form.
$T(n)=aT\left ( \frac{n}{b} \right )\dotplus n^{c}$
Solving this recurrence relation we get,
$T(n)=n^{c}\sum_{k=0}^{log_{b}n-1} \left ( \frac{a}{b^{c}} \right )^{k}\dotplus n^{log_{b}a}$
And geometric sum,
$\sum_{k=0}^{n}x^{n}= \frac{x^{n+1}-1}{x-1}=\Theta \left ( x^{n} \right )$
Now after solving the given relation we get
$T(n)=n^{2}\sum_{k=0}^{log_{3}n-1} \left ( \frac{7}{3^{2}} \right )^{k}\dotplus n^{log_{3}7}$
And the sum of decreasing geometric term $\sum_{k=0}^{log_{3}n-1} \left ( \frac{7}{3^{2}} \right )^{k}$ will be constant.
So $T(n)=n^{2}\dotplus n^{log_{3}7}$
$T(n)=O\left ( n^{2} \right )$