Binomial $(n=7,k=3,p=?) = \text{Geometric}(p=?)$ where Geometric function succeeds on 3rd trial, what is $p$?

Question

I'm trying to find $p$ s.t the Binomial function of it succeeds $3/7$ times and the Geometric function succeeds on the 3rd attempt. I tried equating the functions but it was a mess

$$7C3p^{3}(1-p)^{4}=(1-p)^{2}p => 7C3p^{2}(1-p)^2=1$$

Answer 1

To solve this special quartic equation, take square roots on both sides: $$\sqrt{35}p(1-p)=1$$ Now apply the quadratic formula to solve for $p$.