13.414 Two inspectors set off simultaneously from points $\boldsymbol{A}$ and $B$ to the scene of an incident, point $C$. The first inspector arrived at point $C$ after $a$ minutes. If the second inspector aims to reach point $C$ from point $B$ simultaneously with the first, he would have to spend $c$ minutes less on each kilometer than the first, because the distance from $\boldsymbol{B}$ to $C$ is $\boldsymbol{b}$ km longer than the distance from $\boldsymbol{A}$ to $C$. At what distance from point $A$ did the incident occur?
Solution. Let $x$ be the distance between $A$ and $C$. Then $\dot{x}+b$ is the distance between $B$ and $C$, $\frac{a}{x}$ and $\frac{a}{x+b}$ are the times taken by the first and second inspectors respectively to travel 1 km.

According to the condition, $\frac{a}{x}=\frac{a}{x+b}+c \Rightarrow c x^{2}+b c x-a b=0$.

From this,

Answer: $x=\frac{-b c+\sqrt{b^{2} c^{2}+4 a b c}}{2 c} \text{ km}$.