To find the roots of the quadratic equation

$x^2 - 2x + 3 = 0$

we'll use the quadratic formula:

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

where $a = 1$, $b = -2$, and $c = 3$.

Step 1: Substitute the values into the formula

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Step 2: Simplify

1. Calculate $b^2 - 4ac$: $(-2)^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8$

2. Since the discriminant $-8$ is negative, the roots will be complex numbers.

Step 3: Continue with the formula

$x = \frac{2 \pm \sqrt{-8}}{2}$

Rewrite $\sqrt{-8}$ as $\sqrt{8}i$, where $i$ is the imaginary unit ($i = \sqrt{-1}$).

$x = \frac{2 \pm 2\sqrt{2}i}{2}$

Step 4: Simplify further

$x = 1 \pm \sqrt{2}i$

Final Answer

The roots are:

$x = 1 + \sqrt{2}i \quad \text{and} \quad x = 1 - \sqrt{2}i$

If you have specific answer choices, we can match these roots to the closest option.

---

Would you like further clarification on complex roots or any other detail? Here are some related questions you might explore:

1. What is the discriminant, and how does it determine the nature of roots?
2. How do complex roots appear in the graph of a quadratic equation?
3. What are some applications of complex numbers in engineering or physics?
4. How do complex conjugates relate to each other in quadratic equations?
5. Can all quadratic equations with real coefficients have real roots?

Tip: Complex roots always come in conjugate pairs when the coefficients of the polynomial are real.