The problem asks us to add two complex numbers, $3-2i$ and $4+3i$, and then multiply the result by the complex number $6i$. Finally, we need to write the answer in the form $a+bi$, where $a$ and $b$ are real numbers.

1. Problem Description

The problem asks us to add two complex numbers, $3-2i$ and $4+3i$, and then multiply the result by the complex number $6i$. Finally, we need to write the answer in the form $a+bi$, where $a$ and $b$ are real numbers.

2. Solution Steps

First, add the two complex numbers:

$(3 - 2i) + (4 + 3i)$.

To add complex numbers, we add the real parts and the imaginary parts separately:

$(3+4) + (-2+3)i = 7 + 1i = 7+i$.

Next, multiply the result $(7+i)$ by $6i$:

$(7+i)(6i)$.

To multiply complex numbers, we distribute:

$7(6i) + i(6i) = 42i + 6i^2$.

Recall that $i^2 = -1$. Therefore, we have:

$42i + 6(-1) = 42i - 6 = -6 + 42i$.

Thus, the final result is in the form $a+bi$ with $a=-6$ and $b=42$.

3. Final Answer

$-6 + 42i$