The boat sailed downstream from point A to point B. After an hour’s stay at point B, the boat set off back

The boat sailed downstream from point A to point B. After an hour’s stay at point B, the boat set off back and 5 hours after sailing from point A returned to the same pier. What is the speed of the boat in still water if the distance between points A and B is 40 km and the current speed is 3 km / h? Make an equation corresponding to the condition of the problem, if the letter x denotes the speed of the boat in still water (in km / h).

Let the boat’s own speed be equal to X km / h. At point A and at point B, this speed was the same. All the time on the way back and forth, and also to the parking lot at point B, he spent 5 hours. Then in the movement of everything he was:
5 – 1 = 4 (hours) – the boat was on the way.
The distance between A and B is 40 km, which means that he traveled a distance of 40 km to both point B and point A.
The current speed is 3 km / h. Then, when moving downstream, the speed of the boat is equal to X + 3 km / h, and against the current is equal to X – 3 km / h.
Let’s write down the general formula for finding the path:
S = V * t (where S is the path; V is the speed; t is the time).
Let’s compose and solve the equation:
40 / (X + 3) + 40 / (X – 3) = 4;
40 * (X – 3) + 40 * (X + 3) = 4 * (X ^ 2 – 3 ^ 2);
40X – 120 + 40X + 120 = 4X ^ 2 – 36;
80X = 4X ^ 2 – 36;
-4X ^ 2 + 80X + 36 = 0;
4X ^ 2 – 80X – 36 = 0;
X ^ 2 – 20X – 9 = 0;
By D1:
D1 = (b / 2) ^ 2 – a * c = (-20/2) ^ 2 – (-9) = 10 ^ 2 + 9 = 109;
√D1 = √109 = 10.4.
X1 = (-b / 2 – √D) / a = (10 – 10.4) / 1 = -0.4 – does not satisfy the condition of the problem.
X2 = (-b / 2 + √D) / a = (10 + 10.4) / 1 = 20.4.
Answer: 20.4.