Why guess $\Theta(n^2)$ for the substitution method of worst-case partitioning
With the help from a fellow friend I came to the following conclusion:

We know that we have to make $$n-1,n-2, \dots, 2,1 \text{ comparisons}$$ 

So when looking at the arithmetic series $$\sum_{i=1}^n k = \Theta(n^2)$$

We can conclude that $T(n) = \Theta(n^2)$. 

Please comment if I can improve my answer in any way.