Report sample: https://colab.research.google.com/drive/1XMSfX646Gj-83531zYt1wXZ8RA0N1uOP?usp=sharing

For a simple pendulum

$$ T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow g = 4\pi^2 \frac{L}{T^2}. $$

For a compound pendulum

$$ T = 2\pi \sqrt{\frac{\frac{K^2}{l}+l}{g}} $$

and, hence, a compound pendulum is equivalent to a simple pendulum if

$$ L = \frac{K^2}{l} + l \Rightarrow l^2 - lL + K^2 = 0 $$

which is a quadratic equation with two solutions $l_1$ and $l_2$ where $l_1+l_2=L$ and $l_1l_2=K^2$.

You have to find gravitational acceleration $g$ and radius of gyration $K=\sqrt{l_1l_2}$.