SOLUTION: Thank you in advance for your help. I have attacked this problem many different ways and still come up short. The problem is: The members of a bicycle club rode a 20-mile round

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Question 305501: Thank you in advance for your help. I have attacked this problem many different ways and still come up short. The problem is:
The members of a bicycle club rode a 20-mile round trip route. On the way back, they had a tailwind and averaged 3 mph faster than on the first 10 miles of the trip. If the entire trip took them 1 1/2 hours, what was their average speed on the first half of the ride?
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The members of a bicycle club rode a 20-mile round trip route. On the way back, they had a tailwind and averaged 3 mph faster than on the first 10 miles of the trip. If the entire trip took them 1 1/2 hours, what was their average speed on the first half of the ride?
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d = distance one way = 10 miles
r = rate going without tailwind
d/r + d/(r+3) = 1.5
10/r + 10/(r+3) = 1.5
10(r+3) + 10r = 1.5r(r+3)

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1681 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 12, -1.66666666666667. Here's your graph:

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Ignore the negative solution
r = 12 mph (1st half)