To solve the equation $2(x-1)^2 + 3 = 0$ for $x$ when $y = 0$, follow these steps:

1. Set the equation equal to zero: $2(x-1)^2 + 3 = 0$

2. Isolate the quadratic term: $2(x-1)^2 = -3$

3. Divide both sides by 2: $(x-1)^2 = -\frac{3}{2}$

4. Take the square root of both sides: $x-1 = \pm \sqrt{-\frac{3}{2}}$

   Recall that $\sqrt{-1} = i$, where $i$ is the imaginary unit.

5. Simplify the right-hand side: $x - 1 = \pm i\sqrt{\frac{3}{2}}$

6. Solve for $x$: $x = 1 \pm i\sqrt{\frac{3}{2}}$

So, the solutions are: $x = 1 + i\sqrt{\frac{3}{2}} \quad \text{and} \quad x = 1 - i\sqrt{\frac{3}{2}}$

These are complex roots, meaning there are no real values of $x$ that satisfy the equation.

Would you like further details or explanations on any of these points?

Here are 8 related questions for further exploration:

1. What does it mean for a quadratic equation to have complex roots?
2. How do you graph a function with complex roots?
3. How does the discriminant of a quadratic equation determine the nature of the roots?
4. Can a quadratic equation have one real and one complex root?
5. What are the implications of complex roots for the shape of the parabola?
6. How can you convert a quadratic equation into vertex form?
7. What are some real-world applications of solving quadratic equations with complex roots?
8. How does the imaginary unit $i$ help in solving quadratic equations?

Tip: When encountering complex roots, remember that they always come in conjugate pairs (e.g., $a + bi$ and $a - bi$). This ensures the polynomial's coefficients remain real.