13.422 Watermelons delivered to the base are intended for two stores. The first store immediately began transporting the watermelons and transported them daily in portions of equal mass. The second store began transporting the watermelons $a$ days later and also transported them daily in portions of equal mass, but different from those of the first store. After $b$ days from the start of the transportation operations, half of the initial amount of watermelons remained at the base. How many days did it take to transport all the watermelons from the base, if the transportation ended simultaneously, and the mass of the watermelons received by the first store is equal to the mass of the watermelons received by the second store?
Solution. Let $z$ be the number of days it took to transport all the watermelons from the base, $x$ and $y$ be the daily portions of the transported products by the first and second stores, respectively. Then $x z$ is the mass of watermelons transported by the first store, and $y(z-a)$ by the second store. According to the problem, $x z = y(z-a)$. Additionally, in $b$ days, the amount of products transported is $b x + (b-a) y$, which, according to the problem, equals $\frac{1}{2}(x z + y(z-a))$.

Thus, $\left\{\begin{array}{l}z x=(z-a) y, \\ (z-2 b) x=(2 b-z-a) y\end{array} \Rightarrow \frac{z-a}{z}=\frac{2 b-z-a}{z-2 b} \Rightarrow z^{2}-2 b z+a b=0 \Rightarrow\right.$ $\Rightarrow z=b+\sqrt{b^{2}-a b}$ (the root $z=b-\sqrt{b^{2}-a b}$ is discarded because $z>b$).

Answer: $b+\sqrt{b(b-a)}$.