Bulgaria competition problems, 2022

Consider orthogonal coordinate system and the points $A(1, -2)$ and $B(4, 2)$. Since
$$\begin{aligned} \sqrt{x^2 - 2x + y^2 + 4y + 5} &= \sqrt{(x-1)^2 + (y+2)^2} \\ \sqrt{x^2 - 8x + y^2 - 4y + 20} &= \sqrt{(x-4)^2 + (y-2)^2}, \end{aligned}$$
we conclude that the solutions of the first equation are all points $C(x, y)$ such that $|CA| + |CB| = 5$. But
$$|AB| = \sqrt{(4-1)^2 + (2-(-2))^2} = \sqrt{16+9} = 5 = |CA| + |CB|,$$
and thus the point $C$ lies on the line segment $AB$. The equation of the line $AB$ is $3y = 4x - 10$ so we want to find the solutions of the two equations satisfying $x \in [1, 4]$. Now the solution of the system is given by
$$32x^2 - 80x + 100 = 68 \Leftrightarrow 2x^2 - 5x + 2 = 0 \Leftrightarrow x_1 = \frac{1}{2}, x_2 = 2.$$
Only $x_2$ belongs to the interval $[1, 4]$, thus the only solution is $(2, -\frac{2}{3})$.