This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise F4.
Use Exercise F3 to prove: Any two quadratic extensions of a finite field are isomorphic.
Let $p(x) = x^2+\alpha x+\beta$ be an irreducible polynomial over a finite field $F$. \begin{align*} p(x) &= x^2+\alpha x+\beta \\ &= (x+\frac{\alpha}{2})^2 + (\beta-\frac{\alpha^2}{4}) \end{align*}
Let $\displaystyle y=x+\frac{\alpha}{2}, a = -(\beta-\frac{\alpha^2}{4})$, so $\displaystyle p\left(y-\frac{\alpha}{2}\right) = y^2 - a$. Let $\displaystyle \bar{p}(y) = p\left(y-\frac{\alpha}{2}\right) = y^2 - a$.
Note $a$ must be a non-square. Suppose the opposite. Let $a=c^2$ for some $c\in F$. Then $\displaystyle\bar{p}(y)=p\left(y-\frac{\alpha}{2}\right) = y^2-c^2=(y+c)(y-c)\implies p(x) = \left(x+\frac{\alpha}{2}+c\right)\left(x+\frac{\alpha}{2}-c\right)$, which contradicts that $p(x)$ is irreducible.
Let $\sqrt{a}$ denote the root of $\bar{p}(y)$. In general, $F[x]/\langle p(x + k)\rangle \cong F[x]/\langle p(x)\rangle$ for $k\in F$. Hence, $F(\sqrt{a}) \cong F[y]/\langle \bar{p}(y)\rangle \cong F[y]/\langle p(y-\alpha/2)\rangle\cong F[x]/\langle p(x)\rangle$.
Similar argument can be made so that $F(\sqrt{b})\cong F[x]/\langle q(x)\rangle$ for a different irreducible quadratic polynomial $q(x)$ over $F$ such that $\bar{q}(y) = y^2 - b$.
By Exercise F3, $F(\sqrt{a})\cong F(\sqrt{b})$, where $F(\sqrt{a})$ and $F(\sqrt{b})$ are any two quadratic extensions of a finite field $F$.
Does this look reasonable?