My attempt started as follows. I know that for $f$ to be irreducible, $D = b^2 - 4a$ is not a square in $\Bbb F_p$ (ie $(\frac{D}{p}) = -1$). I also know that $D^{p-1} = 1$, so I see $\sqrt{(D^{p-1})} = 1 \ or \ -1$. I am trying to now see if the fact that $D$ is not a square can give me the desired result $(b^2 - 4a)^{\frac{p-1}{2}} = -1$.
One of my ideas was that $\sqrt{D} \ \in \ \Bbb F_{p^{2}}$. Call $\sqrt{D} = \alpha \in F_{p^{2}}$, then assume $\alpha^{p-1} - 1 = 0$.
I have no idea if this would lead to a contradiction or not (still trying to work it out). I would be super grateful for any comments on if I am heading in the wrong direction or what a next step could be.