After traveling exactly half of its journey from station a to station b, a train was held up for 10 minutes. in order to arrive at city b on schedule, the engine driver had to increase the speed of the train by 12 km/hour. find the original speed of the train before it was held up, if it is known that the distance between the two stations is 120 km.

Let's define variables:
 s = original speed
 s + 12 = faster speed
 The time for the half of the route is:
 60 / s
 The time for the second half of the route is:
 60 / (s + 12)
 The equation for the time of the trip is:
 60 / s + 60 / (s + 12) + 1/6 = 120 / s
 Where,
 1/6: held up for 10 minutes (in hours).
 Rewriting the equation we have:
 6s (60) + s (s + 12) = 60 * 6 (s + 12)
 360s + s ^ 2 + 12s = 360s + 4320
 s ^ 2 + 12s = 4320
 s ^ 2 + 12s - 4320 = 0
 We factor the equation:
 (s + 72) (s-60) = 0
 We take the positive root so that the problem makes physical sense.
 s = 60 Km / h
 
 The original speed of the train before it was held up is:
 s = 60 Km / h

20y5 times 2 is 1010 times -2 is -20, plus y