Finding the minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$ Minimal polynomial of $2+\sqrt3$ over $\mathbb{Q}$.
$(2+\sqrt3)^2$= $7+4\sqrt3$
$\implies$ $((2+\sqrt3)^2-7)^2$=$48$
$\implies$ $((2+\sqrt3)^4-14(2+\sqrt3)^2+1=0$
So $2+\sqrt3$ is a root of $x^4-14x^2+1$. Did i do this correct so far? Any tips on factoring this into quadratics? Thanks!

This strategy can be applied in lots of situations:
\begin{align}
\alpha = 2+\sqrt 3 &\implies \alpha - 2 = \sqrt 3\\
&\implies (\alpha-2)^2 = 3\\
&\implies \alpha^2 - 4\alpha + 1 = 0
\end{align}
so $\alpha$ vanishes on $x^2-4x+1$. It is irreducible since it has no rational roots, so it is the minimal polynomial.
Your $x^4-14x^2+1$ can be factored as $(x^2-4x+1)(x^2+4x+1)$ and it has roots $\pm 2\pm\sqrt 3$. 
To get that factorization, you can first observe that your polynomial has no rational roots (and therefore no linear factors), so it's of the form $(x^2+ax+b)(x^2+cx+d)$. Expanding and comparing coefficients would give you factorization.
