Quadratics Distance-Speed-Time Rate Word Problem I have the following problem from a Textbook

A small plane is travelling between Windsor and Pelee Island (a distance of approx. 60 km) and is directly affected by the prevailing winds. Thus, the actual speed of the plane with respect to the ground is the speed of the plane (160 km/h) plus or minus the wind speed, w. Develop a simplified equation for the total time it takes to make a round trip if the wind speed is w.

I made my best effort to sovle this myself, but in comparing with the textbook answer-key, what I've got and what they've got don't match- and I can't really figure out why.
The textbook answer was:
$$T= \frac{19200}{(160+w)(160-w)}$$
My answer was:
$$T= \frac{120}{(160+w)(160-w)}$$
My logical process for this question was in two steps. I know that Time is going to be distance / speed. In order to determine speed, I need to factor speed with and against the wind, which gives me the quadratic on the bottom. For my distance, I read off the question it is 60km. If it is 60km, and a round trip, the total distance must be 120km.
What am I missing that 120 is transformed into 19200. My initial guesses were a unit issue (i.e metres vs kilometres), but that doesn't make much sense to me. Since they give you the fixed distance, I don't really know what else to do here.
Any and all assistance is appreciated.
 A: $T = \frac {60}{160 + w} + \frac {60}{160-w} = \frac {60(160-w) + 60(160 + w)} {(160 + w)(160 - w)} = \frac {9600 - 60w + 9600 + 60w}{(160 + w)(160 - w)}= \frac {19200}{(160 + w)(160 - w)} = \frac {19200}{25600 +160w-160w-w^2}=\frac {19200}{25600-w^2}$
For this scenario of finding round trip time where you have 2 islands, your wind direction is always parallel to your flight path, and the wind is always in the same direction you could reduce it to this formula.
r=rate (in this instance 160 km/h)
w=wind speed (assuming here is in units km/h)
d=km between the 2 islands
$T = \frac {d(r-w) + d(r +w)}{(r+w)(r-w)} = \frac {dr+dr} {r^2-w^2} = \frac {2dr}{r^2-w^2}$
If we take the problem with no wind.
r= 160 km/h
w=wind speed (assuming here is in units km/h)
d=60km
$\frac {(2)(60)(160)}{160^2-0^2} = \frac {19200}{25600} = \frac {3}{4}$ of an hour
Problem with 30 km/h wind speed
r= 160 km/h
w=30km/h
d=60km
$\frac {(2)(60)(160)}{160^2-30^2} = \frac {19200}{25600-900}= \frac {19200}{24700} = \frac {192}{247}$ of an hour = ~$0.777$ of an hour
A: 
Quite so: the fact that your numerator is off by exactly a factor of  160  should be suspicious.

Thank you for the hint. While the comment above yours directly answered the question, I didn't see it until after I made the connection- and I'm glad about that.
After thinking about why it would be off by a factor of 160, and remembering this is Quadratics, I went back to the question and realized something wasn't being multiplied correctly- or in this case at all.
The final answer is:

This is because there are two separate rates being compared here. The first is the time to get there, and the second is the time to get back- and while it might seem convenient to just stick it all together under one fraction, they are two individual trips with differing speeds, so multiplying them without respect to the other is incorrect.
Thanks to @boojum and @user2661923.
A: $T = \frac{60}{160+w}+\frac{60}{160-w}=\frac{60\times(160-w)}{(160+w)\times(160-w)}+\frac{60\times(160+w)}{(160-w)\times(160+w)}=\frac{9600-60w}{(160+w)\times(160-w)}+\frac{9600+60w}{(160-w)\times(160+w)}=\frac{19200}{(160-w)\times(160+w)}$
