X and Y are independent X~Geom(P) Y~Exp($\lambda $)
Compute the probability that $t^2 + 2\sqrt{x}t + y = 0 $
Steps I've taken so far:
Found where the determinant of the quadratic is $\geq 0$. $\sqrt{4x - 4y)}$ is positive when $x-y \geq 0$
How do I then find the probability that $x-y \geq 0$ has only one real solution?