To solve $\frac{-2}{4 - 3i}$ and write the answer in standard form (as $a + bi$), we need to eliminate the imaginary unit $i$ from the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator.

1. Identify the conjugate: The conjugate of $4 - 3i$ is $4 + 3i$.

2. Multiply both numerator and denominator by the conjugate:

   $\frac{-2}{4 - 3i} \times \frac{4 + 3i}{4 + 3i} = \frac{-2(4 + 3i)}{(4 - 3i)(4 + 3i)}$

3. Expand the numerator:

   $-2(4 + 3i) = -8 - 6i$

   So the numerator becomes $-8 - 6i$.

4. Simplify the denominator using the difference of squares formula:

   $(4 - 3i)(4 + 3i) = 4^2 - (3i)^2 = 16 - 9(-1) = 16 + 9 = 25$

5. Combine:

   $\frac{-8 - 6i}{25} = -\frac{8}{25} - \frac{6}{25}i$

So the answer in standard form is:

$-\frac{8}{25} - \frac{6}{25}i$

Would you like further details on each step, or have any questions?

Here are some related questions to deepen understanding:

1. How do you find the conjugate of a complex number?
2. Why is multiplying by the conjugate helpful in simplifying complex fractions?
3. What does it mean to express a complex number in standard form?
4. How do you add and subtract complex numbers?
5. How do you multiply two complex numbers?

Tip: When dividing by a complex number, always look for its conjugate to simplify the division into real and imaginary parts.