Two cars drove from point A to point B at the same time. The first drove at a constant speed all the way.

Two cars drove from point A to point B at the same time. The first drove at a constant speed all the way. The second traveled the first half of the journey at a speed lower than the first by 14 km / h, and the second half of the journey at a speed of 105 km / h, as a result of which he arrived at B simultaneously with the first car. Find the speed of the first car if it is known to be greater than 50 km / h.

Take distance as 1. Time is equal to distance divided by speed. Let’s take the speed of the first car as x. He traveled all the way at the same speed. So the time he spent is 1 / x. The second car drove half the way at a speed lower than that of the first car by 14 km / h. It turns out that the time he spent on the first half of the journey was 0.5 / (x-14). He traveled the second half of the journey at a speed of 105 km / h. For this half, he spent 0.5 / 105. Cars arrived at the same time means the time spent by the first car is equal to the time of the second one. We summarize 0.5 / (x-14) + 0.5 / 105 = 1 / x. We get a common denominator of 105 (x-14). The first factor is multiplied by 105 and the second by (x-14). We get (0.5 * 105 + 0.5 (x-14)) / (105 (x-14)) = 1 / x; (52.5 + 0.5x-7) / (105x-1470) = 1 / x; We multiply the proportion crosswise. We get 52.5x + 0.5×2-7x = 105x-1470; 0.5×2-59.5x + 1470 = 0. We multiply all the terms by 2 for convenience and get x2-119x + 2940 = 0. Find the discriminant D = b2-4ac = 14161-11760 = 2401. The square root is 49. The first root is (-b + 49) / 2 = (119 + 49) / 2 = 84. The second root is (-b-49) / 2 = (119-49) / 2 = 35. By condition, the vehicle speed is more than 50 km / h.
Answer: 84 km / h speed of the first car.