Image distance for two separated thin lenses ($d_2=\frac{f_2[(L-f_1)d_1-L{f}_1]}{(L-f_1-f_2)d_1+f_1{f}_2-L{f}_1}$)

Use positive distances for a real incident object and a real outgoing image. The first thin lens makes an image at $v_1=f_1{d}_1/(d_1-f_1)$. The second has object distance $L-v_1$, possibly negative, and the thin lens equation gives the displayed expression. At $L=0$ it becomes $d_2=F{d}_1/(d_1-F)$ with $F^{-1}=f_1^{-1}+f_2^{-1}$. For $d_1>0$, a real final image requires both $F>0$ and $d_1>F$; the inequality alone is insufficient when $F<0$.