Yes, that is correct! If the equation is T(n) = 4T(n/2) + n^2, then you can use the Master Theorem to determine the time complexity as follows:

Determine the value of a: a = 4

Determine the value of b: b = 2

Determine the order of f(n): f(n) = n^2, which is O(n^2)

Apply the Master Theorem:

If f(n) = O(n^logb(a)), then T(n) = O(n^logb(a) * log(n)).
If f(n) = O(n^logb(a) * log(n)), then T(n) = O(n^logb(a) * log(n)).
If f(n) = O(n^c) for c > logb(a), then T(n) = O(n^c).
In this case, f(n) = O(n^2), which is greater than logb(a) = log2(4) = 2. Therefore, T(n) = O(n^2 * log(n)).

So the time complexity of this algorithm is O(n^2 * log(n)).
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