jmc

Since $25<30<36$, we have $5<\sqrt{30}<6$. We also know that $7^2=49$, so $\sqrt{50}\approx 7$. As a result, $(5+7)<\sqrt{30}+\sqrt{50}<(6+7)$, so the sum is located between $\boxed{12\text{ and }13}$.

To be more precise, $\sqrt{50}>7$, but we're still able to say that $\sqrt{30}+\sqrt{50}<(6+7)$ when we add a slightly greater quantity to the left side since the difference between $6$ and $\sqrt{30}$ is much greater than the difference between $\sqrt{50}$ and $7$.