need information about this type of distribution I came across a question like this:
"Suppose a large hospital is interested in recruiting patients with a specific medical condition for a experiment. overall, 4 in every 10 patients vesting this hospital suffer from this condition. The physician in charge of the project wants to determine the mean number of patients she needs to examine before she can identify 2 patients with this condition."
Using the line of reasoning for the geometric distribution, and the fact that one of the two patients identified in the experiment must show up in last trial, the other one can be anywhere from first to second from last, I get:
$$P(X=x)=p^2+p(1-p)p+(1-p)p^2+p(1-p)(1-p)p+(1-p)p(1-p)p+p(1-p)(1-p)p^2+...+(1-p)^{x-2}p^2=\sum_{k=2}^x(k-1)(1-p)^{x-2}p^2$$
p - probability of a randomly selected patient will suffer from the condition (p=4/10=0.4)

x - the number of patients the doctor needs to examine before 2 of patients with condition are identified

Is this correct?
What type of distribution is this?

This is the Negative Binomial distribution $\text{NB}(r,p)$, which models the number of failures before you get $r$ successes, where $p=P(\text{Success})$.
In your case, $p=0.4$ and $r=2$. The total number of people she sees $X$ will be a random variable:
$$X = S + 2,\;\;\text{where }S\sim \text{NB}(2,0.4)$$
