The fast train was delayed at the semaphore for 16 minutes. And he eliminated the delay on the stretch of 192 km

The fast train was delayed at the semaphore for 16 minutes. And he eliminated the delay on the stretch of 192 km with a speed exceeding by 10 km / h. put on schedule. Find the speed of the train.

Let us denote by variable x the speed with which the fast train moved in order to eliminate the admitted lag.
Then its planned speed is equal to (x – 10) km / h.
Knowing that in fact he spent 192 km 16 minutes less than the plan, we draw up an equation and determine his real speed:
192 / (x – 10) – 192 / x = 16/60;
192x – 1920x + 1920 = 4x (x -10) / 15;
x ^ 2 – 10x – 7200 = 0;
D = 100 – 4 (-7200) = 1702;
x1 = (10 + 170) / 2 = 90;
x2 = (10 – 170) / 2 = -80.
Answer: The speed of the fast train is 90 km / h.