We are asked to solve the quadratic equation:

\[
x^2 - 2x - 3 = 0
\]

To solve for \( x \), we can use the **quadratic formula**, which is given by:

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

In this case, the equation is in the form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -2 \)
- \( c = -3 \)

Now, plug these values into the quadratic formula:

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

Simplify the terms:

\[
x = \frac{2 \pm \sqrt{4 + 12}}{2}
\]
\[
x = \frac{2 \pm \sqrt{16}}{2}
\]
\[
x = \frac{2 \pm 4}{2}
\]

This gives two possible solutions:

1. \( x = \frac{2 + 4}{2} = \frac{6}{2} = 3 \)
2. \( x = \frac{2 - 4}{2} = \frac{-2}{2} = -1 \)

So the solutions are:

\[
x = 3 \quad \text{or} \quad x = -1
\]

### Do you want a detailed explanation of any of the steps, or have further questions?

Here are 5 related questions you might find interesting:
1. What if the quadratic equation had no real solutions? 
2. How can you factor the quadratic equation \(x^2 - 2x - 3 = 0\) instead of using the quadratic formula?
3. What is the discriminant and how does it help in determining the number of solutions to a quadratic equation?
4. Can a quadratic equation have complex solutions? How?
5. What are some common methods for solving quadratic equations beyond the quadratic formula?

**Tip**: Always check the discriminant (\(b^2 - 4ac\)) to determine whether your quadratic equation has real or complex solutions.

Learn how to solve the quadratic equation x^2 - 2x - 3 = 0 using the quadratic formula. The step-by-step solution includes simplifying the equation and finding the two possible solutions: x = 3 or x = -1. Ideal for students in Grades 7-9.

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