SOLUTION: A university flew its football team round trip for a game with a rival team. The distance was 540 miles each way, and the round trip required a total flying time of 8.8 hours. Beca

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Question 183980: A university flew its football team round trip for a game with a rival team. The distance was 540 miles each way, and the round trip required a total flying time of 8.8 hours. Because of wind, the average speed of the plane in one direction was 30 miles per hour faster than it was in the opposite direction. What would be the speed of the plane in still air?
Answer by checkley77(12569) About Me  (Show Source):
You can put this solution on YOUR website!
540/(x+15)+540/(x-15)=8.8
[540(x-15)+540(x+15)]/(x+15)(x-15)=8.8
[540x-8100+540x+8100]/(x^2-225)=8.8
1080x/(x^2-225)=8.8
1080x=8.8x^2-1980
8.8x^2-1080x-1980=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x=(1080+-sqrt[-1080^2-4*8.8*-1980])/2*8.8
x=(1080+-sqrt[1,166,400+69,696]/17.6
x=(1080+-sqrt1,236,096)/17.6
x=(1080+-1,111.7985)/17.6
x=(1080+1,111.7985)/17.6
x=2,191.7985/17.6
x=124.53 mph. is the epeed of the plane in still air.