To solve $(-6 - i)(3 - 2i)$, we will use the distributive property (also known as FOIL for binomials), multiplying each term in the first complex number by each term in the second complex number.

1. First, multiply the real parts: $-6 \times 3 = -18$

2. Then, multiply the real part of the first number by the imaginary part of the second number: $-6 \times (-2i) = 12i$

3. Next, multiply the imaginary part of the first number by the real part of the second number: $-i \times 3 = -3i$

4. Finally, multiply the imaginary parts: $-i \times (-2i) = 2i^2$ Since $i^2 = -1$, this becomes: $2i^2 = 2(-1) = -2$

Now, combine all the results: $-18 + 12i - 3i - 2 = (-18 - 2) + (12i - 3i) = -20 + 9i$

So, $(-6 - i)(3 - 2i) = -20 + 9i$.

Would you like more details or further clarification?

Here are five related questions for you:

1. How do you multiply complex numbers in general?
2. What happens when you multiply two purely imaginary numbers?
3. How do you convert a complex number from rectangular to polar form?
4. What is the significance of the imaginary unit $i$ in mathematics?
5. How do you find the conjugate of a complex number?

Tip: When multiplying complex numbers, always remember to simplify $i^2 = -1$ to handle the imaginary components correctly.