From point A to point B, the distance between which is 20 km, a pedestrian left. Simultaneously with it, a cyclist left

From point A to point B, the distance between which is 20 km, a pedestrian left. Simultaneously with it, a cyclist left point B to point A and met a pedestrian 50 minutes after leaving B. How long would it take for a pedestrian to go all the way from A to B if it is known that the cyclist would have done the same the way is 4 hours faster than a pedestrian?

Given:
S = S1 + S2 = 20 km
S1 = vp * Tv
S2 = vw * Tw
TV = 50 min = 5/6 h – time before meeting
(5/6) * vp + (5/6) * vv = 20
5 * (vp + vv) = 120
vp + vv = 24
vv = 24-vp
also from the conditions of the assignment it is known:
tp = tv + 4
S = vp * tp
S = vw * tv = vw * (tp-4)
vp * tp = vw * (tp-4)
earlier it was found out vv = 24-vp, therefore
vp * tp = (24-vp) * (tp-4)
vp = S / tp = 20 / tp, then we solve the following expression:
20 / tp * tp = (24-20 / tp) * (tp-4)
20 = 24 * tp-20-96 + 80 / tp
0 = 24 * tp-136 + 80 / tp
80 / tp = 136-24 * tp
80 = 136 * tp-24 * tp ^ 2
10 = 17 * tp-3 * tp ^ 2
3 * tp ^ 2-17 * tp + 10 = 0
we solve the quadratic equation:
D = (- 17) ^ 2-4 * 3 * 10 = 169
tp1 = – (- 17) + √169 / 2 * 3 = (17 + 13) / 6 = 5h
tp2 = – (- 17) -√169 / 2 * 3 = (17-13) / 6 = 2 / 3h
Of the two solutions, the 5 o’clock answer seems to be more appropriate, since from the conditions of the problem, a cyclist travels a given distance 4 hours faster than a pedestrian, and the time cannot be negative.
Answer: 5 hours