A train normally travels 240 km at a certain speed. One day, due to bad weather, the train's speed is reduced by 20 km/h so that
the journey takes two hours longer. Find the normal speed

Normal-travel DATA:

distance = 240 km

rate = x km/h

time = d/r

= 240/x hrs

_______________________

Bad weather DATA:

distance = 240 km

rate = (x-20) km/h

time = 240/(x-20) hrs

______________________

Bad time - normal time = 2 hrs

240/(x-20) - 240/x = 2

120/(x-20) - 120/x = 1

120x -120(x-20) = x(x-20)

20 = x^2 -20x

x^2 - 20x - 20 = 0

x = [20 +- sqrt(20^2 - 4*1*-20)]/2

x = [20 +- sqrt(480)]/2

x = [20 +- 4sqrt(30)]/2

x = 10 +- 2sqrt(30)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x