many thanks! I would try to correct the question

Marco: any smooth rational curve on K3 has square -2. Conversely, using Riemann-Roch, you can prove that any line bundle $L$ with $c_1^2=-2$ satisfies $H^2(L)\neq 0$ or $H^2(L^*)'\neq 0$, and the zero divisor of its section contains a smooth rational curve as a component

Samir: many thanks, this is what I need I think.

Jason: for McMullen's nonprojective K3 surfaces, the Picard lattice is either degenerate or negative definite, and such lattices are easy to construct in any rank.

the question is, for which $d$ we can be sure there are no -2 vectors? What about $d=12$? What about $d=20$?

sorry, i misread the question

they are both represented by $\partial \bar\partial f$ (up to a sign and a constant)