To solve the quadratic equation $x^2 - (3\sqrt{2} + 2i) \cdot x + 6\sqrt{2} = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 1$, $b = -(3\sqrt{2} + 2i)$, and $c = 6\sqrt{2}$.

1. Calculate the discriminant: $\Delta = b^2 - 4ac$ $\Delta = \left(-(3\sqrt{2} + 2i)\right)^2 - 4(1)(6\sqrt{2})$

2. Simplify $b^2$: $b^2 = (3\sqrt{2} + 2i)^2 = (3\sqrt{2})^2 + 2(3\sqrt{2})(2i) + (2i)^2$ $= 18 + 12\sqrt{2}i - 4 = 14 + 12\sqrt{2}i$

3. Simplify the discriminant: $\Delta = 14 + 12\sqrt{2}i - 24\sqrt{2}$ $\Delta = 14 - 24\sqrt{2} + 12\sqrt{2}i$

4. Find the roots using the quadratic formula: $x = \frac{3\sqrt{2} + 2i \pm \sqrt{14 - 24\sqrt{2} + 12\sqrt{2}i}}{2}$

Since the discriminant is complex, solving for the roots involves further simplifying the expression under the square root.

Would you like to continue with the detailed steps for finding the roots, or do you have any specific questions so far?

Here are some related questions to consider:

1. What is the significance of the discriminant in determining the nature of the roots?
2. How do you handle complex numbers in the quadratic formula?
3. What does it mean when the discriminant is complex?
4. How do you simplify complex square roots?
5. Can this equation have real solutions?

Tip: When solving quadratic equations with complex coefficients, it's important to carefully manage the arithmetic of complex numbers to avoid errors.