Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 463: 66

$\dfrac{3+2i}{7}$

Using $i^2=-1$, the expression $ \dfrac{2-3i}{-7i} $ simplifies to \begin{array}{l} \dfrac{2-3i}{-7i} \cdot \dfrac{i}{i} \\\\= \dfrac{2i-3i^2}{-7i^2} \\\\= \dfrac{2i-3(-1)}{-7(-1)} \\\\= \dfrac{3+2i}{7} .\end{array}