First, we need to solve it as:
[tex](-2\sqrt{5}-3i)^2=(-2\sqrt{5})^2-2\cdot(-2\sqrt{5})\cdot3i+(3i)^2[/tex]Because:
[tex](a-b)^2=a^2-2\cdot a\cdot b+b^2[/tex]Additionally:
[tex]i^2=-1[/tex]So, the initial expression is equal to:
[tex]\begin{gathered} (-2\sqrt{5}-3i)^2=(-2\sqrt{5})^2-(2\cdot(-2\sqrt{5})\cdot3i)+(3i)^2 \\ (-2\sqrt{5}-3i)^2=(-2)^2\cdot(\sqrt{5})^2+12\sqrt{5}i+(3^2\cdot i^2) \\ (-2\sqrt{5}-3i)^2=4\cdot5+12\sqrt{5}i+9\cdot(-1) \\ (-2\sqrt{5}-3i)^2=20+12\sqrt{5}i-9 \\ (-2\sqrt{5}-3i)^2=11+12\sqrt{5}i \end{gathered}[/tex]Answer: 11 + (12√5)i
