Prove that every finite extension field of $\Bbb R$ is either $\Bbb R$ itself or isomorphic to $\Bbb C$.
I tried in this way.Let $E$ be a a finite extension of $\Bbb R$. Then $E$ is an algebraic extension of $\Bbb R$. Let $\alpha \in E-R$ .then $\exists p(x)\in \Bbb R[x]$ such that $p(\alpha)=0$. Surely $\deg(p(x)=2$ since otherwise it would be reducible.
Now $\Bbb R(\alpha)\cong \Bbb R[x]/<p(x)>$. Now $[\Bbb R(\alpha):\Bbb R]=2$ .Also $[\Bbb C:\Bbb R]=2\implies \Bbb R(\alpha)\cong \Bbb C$.
Since $\Bbb C$ is algebraically closed so is $\Bbb R(\alpha)$ and so it has no proper algebraic extensions.
Can I conclude from here that $E=\Bbb C$?
Please suggest steps if needed.I would be very grateful
You have already correctly deduced that the polynomial $p$ is of degree $2$ if $p$ is irreducible. Therefore we have $$[\mathbb{R}(\alpha) : \mathbb{R}] = 2.$$ Moreover, since $\mathbb{C}$ is the algebraic closure of $\mathbb{R}$ there is an embedding from $\mathbb{R}(\alpha)$ to $\mathbb{C}$ and we can identify it with a subset, i.e. $\mathbb{R}(\alpha) \subseteq \mathbb{C}$. Now we have $$2 = [\mathbb{C} : \mathbb{R}] = [\mathbb{C} : \mathbb{R}(\alpha)] \cdot [\mathbb{R}(\alpha) : \mathbb{R}] = [\mathbb{C} : \mathbb{R}(\alpha)] \cdot 2.$$ We deduce $[\mathbb{C} : \mathbb{R}(\alpha)] = 1$ and thus $\mathbb{C} = \mathbb{R}(\alpha)$.