Two cars leave points A and B at the same time and meet at 12 noon. If the speed of the first is doubled

Two cars leave points A and B at the same time and meet at 12 noon. If the speed of the first is doubled, and the speed of the second is left the original, then the meeting will occur 56 minutes earlier. If the speed of the second is doubled, and the speed of the first remains the same, then they will meet 65 minutes earlier. Determine the time of the meeting in the case when the speed of both would be doubled.

Legend:
S is the distance;
V 1 – the speed of the first car;
V 2 – the speed of the second car;
T is time.
S = (V 1 + V 2) * T;
S = (2 * V 1 + V 2) * (T – 14/15);
S = (V 1 + 2 * V 2) * (T – 13/12).
It turns out:
3 / T = 1 / (T – 14/15) + 1 / (T – 13/12).
30 * T ^ (2) – 121 * T + 91 = 0
D = 14641 – 4 * 30 * 91 = 3721;
T 1 = 1;
T 2 = 182/60.
T = S / (2 * (V 1 + V 2)) = T 2/2 = 1 hour 31 minutes.
12 hours – 1 hour 31 minutes = 10 hours 29 minutes.
Answer: the meeting time for two motorists is 10 hours 29 minutes.