To write a quadratic equation with roots $3 + 2i$ and $3 - 2i$, we can use the fact that if $a$ and $b$ are roots of a quadratic equation, then the equation can be written as:

$(x - a)(x - b) = 0$

Given roots $3 + 2i$ and $3 - 2i$, we can substitute them as $a$ and $b$:

$(x - (3 + 2i))(x - (3 - 2i)) = 0$

Expanding this expression:

1. Rewrite the terms as a difference of squares:

   $= (x - 3 - 2i)(x - 3 + 2i)$

2. Use the identity $(A - B)(A + B) = A^2 - B^2$, where $A = x - 3$ and $B = 2i$:

   $= (x - 3)^2 - (2i)^2$

3. Now, expand $(x - 3)^2$ and $(2i)^2$:

   $= (x^2 - 6x + 9) - (4i^2)$

4. Since $i^2 = -1$, $4i^2 = 4(-1) = -4$:

   $= x^2 - 6x + 9 + 4$

5. Simplify:

   $= x^2 - 6x + 13$

Therefore, the quadratic equation with roots $3 + 2i$ and $3 - 2i$ is:

$x^2 - 6x + 13 = 0$

Would you like more details on any step, or do you have questions?

Here are some related questions to explore:

1. How can complex roots be identified in a quadratic equation?
2. What is the relationship between the coefficients and the roots of a quadratic equation?
3. Can a quadratic equation have more than two complex roots?
4. How do we derive the discriminant for equations with complex roots?
5. What real-world applications involve quadratic equations with complex roots?

Tip: For a quadratic equation with complex roots, the imaginary parts of the roots will always cancel each other out, leading to a quadratic equation with real coefficients.