The equation you've provided is:

\[
2x^2 - 3x - 5 = 0
\]

To solve this quadratic equation, we can use the quadratic formula:

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

Here, the coefficients are:
- \(a = 2\)
- \(b = -3\)
- \(c = -5\)

Substituting these values into the quadratic formula:

\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}
\]

Simplifying step by step:

\[
x = \frac{3 \pm \sqrt{9 + 40}}{4}
\]
\[
x = \frac{3 \pm \sqrt{49}}{4}
\]
\[
x = \frac{3 \pm 7}{4}
\]

This gives two solutions:

1. \(x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5\)
2. \(x = \frac{3 - 7}{4} = \frac{-4}{4} = -1\)

So, the solutions are:

\[
x = 2.5 \quad \text{or} \quad x = -1
\]

Would you like more details on any of these steps or have any other questions?

---

Here are some related questions:
1. How do you derive the quadratic formula?
2. What are some alternative methods to solve quadratic equations?
3. How do you determine the discriminant and what does it tell you about the solutions?
4. Can quadratic equations have complex solutions?
5. How can quadratic equations be applied to real-world problems?

**Tip:** Always check the discriminant (\(b^2 - 4ac\)) before solving a quadratic equation; it can tell you how many real solutions to expect.

Learn how to solve the quadratic equation 2x^2 - 3x - 5 = 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, including the quadratic formula and solving quadratic equations. Suitable for students in Grades 9-12.

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