A friend of mine gave me this problem:
A man who walks at a constant speed goes to his barn 30 miles away with a 2 mph wind pushing against him. After arriving at the barn he remembers he forgot something, so he walks back, but this time the 2 mph wind is working in his favor. When he reaches his house, the entire trip so far has taken 8 hours. What is the man's normal walking speed (i.e. if there is no wind)?
I set it up this way:
$30 = (r-2)t_1 = (r+2)t_2$ where $t_1 + t_2 = 8$
$30/(r-2) + 30/(r+2) = 8$
$30(r+2) + 30(r-2) = 8(r+2)(r-2)$
$2r^2 - 15r -8 = 0$
$r = -1/2, 8 \implies r = 8$
Did I overlook something really obvious or did this actually require the quadratic formula to solve?
You are assuming a $2$ mph wind changes his walking speed by $2$ mph. That seems unreasonable to me. I would hang my hat on the constant speed and say he covered $60$ miles in $8$ hours, so walked $7.5$ mph. The way you read the problem you have solved it correctly. He walks $6$ mph outbound for $5$ hours and $10$ mph inbound for $3$ hours. In any case he walks very fast.