1) Writing down the other root:

Since the quadratic equation has real coefficients, if one of the roots is a complex number, its conjugate must also be a root. The conjugate of $6 - 2i$ is $6 + 2i$. Therefore, the other root is:

$6 + 2i$

2) Finding the quadratic equation:

Let the roots of the quadratic equation be $z_1 = 6 - 2i$ and $z_2 = 6 + 2i$. The general form of a quadratic equation with roots $z_1$ and $z_2$ is:

$(z - z_1)(z - z_2) = 0$

Substitute the values of $z_1$ and $z_2$:

$(z - (6 - 2i))(z - (6 + 2i)) = 0$

This is a difference of squares, and can be simplified as follows:

$(z - 6 + 2i)(z - 6 - 2i) = (z - 6)^2 - (2i)^2 = (z - 6)^2 + 4 = 0$

Now expand $(z - 6)^2$:

$(z - 6)^2 = z^2 - 12z + 36$

Thus, the quadratic equation becomes:

$z^2 - 12z + 36 + 4 = 0$

Simplifying further:

$z^2 - 12z + 40 = 0$

Final answer:

The quadratic equation is:

$z^2 - 12z + 40 = 0$

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Would you like further details or explanations on any of the steps? Here are some related questions for further exploration:

1. How do you find the conjugate of a complex number?
2. What is the discriminant of this quadratic equation, and what does it indicate?
3. How can you factorize quadratic equations with real coefficients?
4. How are complex roots represented on the complex plane?
5. Can all quadratic equations with complex roots be written in terms of real coefficients?

Tip: When working with quadratic equations, always check if the discriminant gives real or complex roots by calculating $b^2 - 4ac$.