Aboat traveled downstream a distance of 80 mi and then came right back. if the speed of the current was 8 mph and the total trip took 5 hours and 20 minutes, find the average speed of the boat relative to the water.

Velocity of the boat relative to the stream = b

Velocity ot the boat downstream = b + 8

Velocity of the boat upstream = b - 8

v = d / t => t = d / v

Time downstream = 80 /  [b + 8]

Time upstream = 80 / [b - 8]

Total time 80 / [b + 8] + 80 / [b - 8] = 17/3 ... [ I converted 5 h + 20 m to 1/

 hours]

Now you have to solve

 80 / [b + 8] + 80 / [b - 8] = 17/3

80[b - 8] + 80[b + 8] = [17/3](b+8)(b-8)

Divide by 80

b - 8 + b + 8 = [17/(80*3)] (b^2 - 64)

2b = [17/240]b^2 - 64*17/240

0.0708b^2 -2b - 4.53333 = 0

Use the quadratic formula to solve. You will obtain b = - 2.11 and b = 30.36

Only the positve result make sense. Then the answer is b = 30.36 mph.

Let's verify that that result is coherent:

Downstream velocity = 30.36 + 8 = 38.36 mi / h
Downstream time = 80mi / 38.36 mi/h = 2.085 h

Upstream velocity = 30.36 - 8 = 22.36 mi / h
Upstream time = 80 mi / 22.36 mi/h = 3.578 h

Total time = 2.085 h + 3.578 h = 5.663 h, which is 5 h + 20 min; then, the result is correct.
 

It's have to be 12 or idk