Task B-2.4. The glider had to return to a place it had passed to refuel at the port. It was delayed for 12 minutes. This delay was compensated for on the remaining 60 km of the journey by increasing the speed by 10 km/h. What was the glider's speed before docking at the port?
## Solution.

Let $x \mathrm{~km} / \mathrm{h}$ be the speed of the glider before it docked in the harbor to refuel. With this speed, the remaining part of the journey, $60 \mathrm{~km}$, would take $\frac{60}{x}$ hours, and with the increased speed, it would take $\frac{60}{x+10}$ hours. The difference in these times must be equal to the time spent in the harbor, i.e.,

$$
\frac{60}{x}-\frac{60}{x+10}=\frac{1}{5}
$$

From this, we get $300(x+10)-300 x=x(x+10)$, or

$$
x^{2}+10 x-3000=0
$$

The solutions to this equation are $x_{1}=-60$ and $x_{2}=50$.

Before docking in the harbor, the glider was traveling at a speed of $x=50 \mathrm{~km} / \mathrm{h}$.