13.134. According to the program, two machines on the assembly line should process the same number of parts in $a$ hours. The first machine completed the task. The second machine turned out to be not quite serviceable, worked with interruptions, as a result of which it processed $n$ fewer parts than the first in the same time. On average, it took $b$ minutes longer to process one part on the second machine than on the first. How many parts did each machine process
Solution.

Let $x$ be the number of parts processed by the first machine, and $x-n$ be the number of parts processed by the second machine. The time spent on processing one part on the first machine is $\frac{a}{x}$ hours, and on the second machine it is $-\frac{a}{x-n}$ hours. According to the condition, $\frac{a}{x-n}=\frac{a}{x}+\frac{b}{60}$, from which we get

\[
\begin{aligned}
& x=\frac{b n+\sqrt{b^{2} n^{2}+240 a b n}}{2 b} ; x-n=\frac{-b n+\sqrt{b^{2} n^{2}+240 a b n}}{2 b} \\
& \text { Answer: } \frac{b n+\sqrt{b^{2} n^{2}+240 a b n}}{2 b} \text { and } \frac{-b n+\sqrt{b^{2} n^{2}+240 a b n}}{2 b} \text { parts. }
\end{aligned}
\]