# initial value problem, mix problem

A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the tank at 6 gallons per minute and the well-stirred solution runs out at 5 gallons per minute. How long will it be until there are 25 pounds of salt in the tank?

I set up the differential equation as $dQ/dt=\frac{6-5Q}{(50+t)}$.

Then I used the integrating factor method to ultimately come out with $Q=\frac{t+50+C}{(t+50)^5}$. When I plug in the initial value $Q(0)=30$, I get an outrageously high number for $C$, which may be correct, but it asks how long it will take until there are 25 pounds of salt left in the tank, so logically I plug in 25 to $Q$ and solve for $t$ to find the time it will take but I keep coming out with a negative number, which makes no sense.

The amount change of brine $V(t)=(6-5)t=t$;
Amount of salt in equals $0$ since there is NO SALT in;
Amount of salt out equals $\frac{5Q}{V(t)}=\frac{5Q}{(50+t)}$;
Then try $dQ/dt$ = In-Out.