Answer
$f_{2}= \displaystyle \frac{ff_{1}}{f_{1}-f}, \quad $focal length of glasses
Work Step by Step
$f=\displaystyle \frac{f_{1}f_{2}}{f_{1}+f_{2}}$
... multiply with the LCD, $(f_{1}+f_{2})$
$f(f_{1}+f_{2})=f_{1}f_{2}$
$ff_{1}+ff_{2}=f_{1}f_{2}$
... transfer $f_{1}f_{2}$ to the LHS: add $-f_{1}f_{2}-ff_{1}$
$ ff_{2}-f_{1}f_{2}=-ff_{1}\quad$ ... factor out $f_{2}$
$ f_{2}(f-f_{1})=-ff_{1}\quad$ ... divide with $(f-f_{1})$
$f_{2}=\displaystyle \frac{-ff_{1}}{f-f_{1}}$
$f_{2}= \displaystyle \frac{ff_{1}}{f_{1}-f}$
(see example 5)
focal length of glasses,
$f_{1}=$ distance from the lens
$f_{2}=$ distance from the lens to your retina
