Why is every smooth quartic in $\Bbb{P}^3$ a K3 surface? Usually the first example of a K3 surface presented to us is the Fermat quartic $x_0^4+x_1^4+x_2^4+x_3^4=0$ in $\Bbb{P}_\Bbb{C}^3$. 
But I've just found out that actually any smooth quartic in $\Bbb{P}^3$ is K3, and I'm trying to understand why.
I know that since $S\subset\Bbb{P}^3$ is a smooth quartic, then  in terms of linear equivalence we have $S\sim 4H$, where $H\subset \Bbb{P}^3$ is a hyperplane. By the adjunction formula:
$$K_S=(K_{\Bbb{P}
^3}+S)\big|_S\sim(-4H+4H)\big|_S=0$$
Now, to prove that $h^1(S,\mathcal{O}_S)=0$, I have no idea what to do.
I imagine that the Fermat quartic has nothing special, and we should be able to prove $h^1=0$ for any smooth quartic without additional difficulties. But I really don't know how to do that.
 A: First note that we have the short exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^3}(-4) \rightarrow \mathscr{O}_{\mathbb{P}^3} \rightarrow \mathscr{O}_S \rightarrow 0.$$ You get what you want then by considering the long exact sequence of cohomology groups plus the fact that $H^1(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}) = H^2(\mathbb{P}^3,\mathscr{O}_{\mathbb{P}^3}(-4)) = 0$.
Edit: 
Since you were asking for some references, I would recommend these lecture notes or their newer (and different) version. In general I can only recommend the two lecture notes full notes (older notes) and very recent full notes by Gathmann. I think everything is well motivated, explained and contains detailed examples etc. He will also explain all the things you need to understand my answer, i.e. the long exact cohomology sequence associated to a short exact sequence and the computation of the cohomology of the twisted sheaves on projective spaces.
A: Use Hodge theory $$H^1(S,\mathcal{O})=H^{0,1}(S)\subset H^1(S,\mathbb C),$$
together with Lefschetz hyperplane theorem $$H^1(S,\mathbb Z)\cong H^1(\mathbb P^3,\mathbb Z)=0.$$
