The discharge rate of the outlet pipe is X and the input rate is Y. If the volume of the pool is V then X=V/T1 where T1 is the time it takes to empty the pool with volume V. To fill the same volume T2=V/Y. T1=T2+2.
When both pipes are open and working, the effective rate of fill=Y-X and T3=8 hours. Y-X=V/T3=V/8.
Therefore, since V=YT2=XT1=X(T2+2) we have Y=V/T2 and X=V/(T2+2), so Y-X=V(1/T2-1/(T2+2))=V/8.
Cancel out V: 1/T2-1/(T2+2)=1/8. Multiply through by 8T2(T2+2): 8T2+16-8T2=T2^2+2T2.
T2^2+2T2-16=0. Complete the square: T2^2+2T2+1=17; (T2+1)^2=17, so T2+1=√17 (T2 has to be positive), making T2=√17-1=3.12 hrs. T1=5.12 hrs, 2 hours longer.
CHECK
1/3.12-1/5.12=1/8 (approx). To the nearest tenth of an hour the time to fill the pool would be T2=3.1 hours.