Suppose I have a chest. When you open the chest, there is a 60% chance of getting a prize and a 40% chance of getting 2 more chests. Let $X$ be the number of prizes you get. What is its variance?

Computing $E[X]$ is fairly straight forward: $E[X] = .4 \cdot 2 \cdot E[X] + .6$ which leads to $E[X] = 3$, but I'd also like to know the variance of the number of prizes, not just the average. $Var[X] = E[X^2] - E[X]^2 = E[X^2] - 9$, but I'm having trouble with $E[X^2]$. Anyone have any idea if this is simple? From simulation, I know that the variance is ~30.

Thanks