proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer 
Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
  an algebraic integer?

An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bbb Q({\sqrt{6}})=\Bbb{Q}(\alpha)/(\alpha^2-6)$, and with that:
$$\beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies \beta= \frac{(3+2\sqrt{6})(1+\sqrt6)}{-5}\implies -5\beta =15+5\sqrt{6}\implies (-5\beta-15)^2=25(6)\implies 5^2\beta^2+2(15)(5)\beta+15^2-5^2(6)=0,$$
and thus $5^2x^2+6(5)^2x+5^2(3)\in\Bbb{Z}[x]$ is the desired polynomial.
Is this correct?

$$ \beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies $$
$$\beta= \frac{(3+2\sqrt{6})(1+\sqrt 6)}{-5}\implies $$
$$-5\beta =15+5\sqrt{6}\implies $$
$$ \beta =-3-\sqrt 6$$  
$$ \beta ^2 + 6\beta +3=0 $$
