This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise F5.
If $a$ and $b$ are nonsquares in $\Bbb{R}, a/b$ is a square (why?). Use the same argument as in Exercise F4 to prove that any two simple extensions of $\Bbb{R}$ are isomorphic (hence isomorphic to $\Bbb{C}$).
Obviously any positive real number is a square and any negative real number is a non-square in $\Bbb{R}$. If $a,b$ are non-squares in $\Bbb{R}$, $a/b$ must be positive and therefore a square.
Using the same argument in Exercise F4, we can find an irreducible quadratic polynomial $p(x)$ over $\Bbb{R}$, such that $\bar{p}(y) = y^2 -a$. Let $\sqrt{a}$ denotes the root of $\bar{p}(y)$. Then $\Bbb{R}(\sqrt{a})\cong\Bbb{R}[y]/\langle\bar{p}(y)\rangle$. Similarly, $\Bbb{R}(\sqrt{b})\cong\Bbb{R}[y]/\langle\bar{q}(y)\rangle$.
Using the same argument in Exercise F3, $\Bbb{R}(\sqrt{a})\cong\Bbb{R}(\sqrt{b})$. This implies any two simple extensions of $\Bbb{R}$ are isomorphic, and hence isomorphic to $\Bbb{R}(\sqrt{-1})=\Bbb{C}$.
Correct?