30. On the x-axis, find a point that is 5 units away from the point $M(1 ; 3)$.
Solution. Let the required point be denoted as \( A(x ; 0) \) (since by the condition it lies on the x-axis). Then the length of the segment \( AM \) can be expressed by the formula \( |AM| = \sqrt{(x_M - x_A)^2 + (y_M - y_A)^2} \), from which, substituting the coordinates of the points and the known distance, we have

\[
5 = \sqrt{(1 - x)^2 + (3 - 0)^2}
\]

We will square both sides of this equation, then expand the brackets and combine like terms:

\[
\begin{gathered}
25 = (1 - x)^2 + 3^2 ; 25 = 1 - 2x + x^2 + 9 ; 25 = x^2 - 2x + 10 \text{ or } \\
x^2 - 2x - 15 = 0 .
\end{gathered}
\]

We will solve the quadratic equation:

\[
\begin{gathered}
D = b^2 - 4ac = 4 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 ; \sqrt{D} = 8 \\
x_{1,2} = \frac{-b \pm \sqrt{D}}{2a} ; x_1 = \frac{2 - 8}{2} = -3 ; x_2 = \frac{2 + 8}{2} = 5
\end{gathered}
\]

Thus, we obtain two points: \( A_1(-3 ; 0), A_2(5 ; 0) \) (Fig. 39).