= Solution
The roots are $\pm\sqrt2$ and $\pm\sqrt3$, so the <splitting field> is $K=\mathbb Q(\sqrt2,\sqrt3)$. The first adjunction has <degree of a field extension>[degree] $2$. To see that the second does too, suppose $\sqrt3=u+v\sqrt2$ with $u,v\in\mathbb Q$. Squaring and comparing rational and irrational parts gives $uv=0$ and $u^2+2v^2=3$. If $v=0$, then $u^2=3$; if $u=0$, then $v^2=3/2$. Neither is a rational square, as follows by comparing prime exponents in numerator and denominator. Thus $\sqrt3\notin\mathbb Q(\sqrt2)$. The <tower law> now yields
$$
\boxed{[K:\mathbb Q]=4,\qquad K=\mathbb Q(\sqrt2,\sqrt3).}
$$
A <basis> is $1,\sqrt2,\sqrt3,\sqrt6$.
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