SOLUTION: A pilot flies 500 miles against a 20-mile-per-hour wind. On the next day, the pilot flies back home with a 10-mile-per-hour tail wind. The total trip (both ways) takes 4 hours.

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Question 428032: A pilot flies 500 miles against a
20-mile-per-hour wind. On the next day, the pilot
flies back home with a 10-mile-per-hour tail wind.
The total trip (both ways) takes 4 hours. Find the
speed of the airplane without a wind.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
plane speed in still air x
wind speed 20

against wind x- 10
with wind x+10

Distance= 500 miles

Time agianst + time with wind =4 hours
t=d/r
500/(x+20)+500/(x-20)= 4
LCD = (x-20)(x +20)
500*(x-20)+500(x+20)=4
500x-10000+500x+10000=4x ^2-400
1000x=4x ^2-400
4x^2-1000x-400=0
Find the roots of the equation by quadratic formula
a= 4 b= -1000 c= -400
b^2-4ac= 1000000 - 6400
b^2-4ac= 1006400
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=(1000+1003.19 )/8
x1= 250.4 mph
x2=(1000-1003.19) /8
x2=-0.4
Ignore negative value
250.4 mph =speed in still air