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.

Learn how to solve the equation 2(x-1)^2 + 3 = 0 for complex roots using the quadratic formula and understand the implications of complex numbers in quadratic equations. Suitable for advanced high school students and those studying complex numbers.

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