To solve the quadratic equation \( 2x^2 + 3x - 9 = 0 \), we can use the **quadratic formula**:

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

For the equation \( 2x^2 + 3x - 9 = 0 \), the coefficients are:
- \( a = 2 \)
- \( b = 3 \)
- \( c = -9 \)

Now, let's substitute these values into the quadratic formula:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-9)}}{2(2)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 72}}{4}
\]
\[
x = \frac{-3 \pm \sqrt{81}}{4}
\]
\[
x = \frac{-3 \pm 9}{4}
\]

So, we have two possible solutions:
1. \( x_1 = \frac{-3 + 9}{4} = \frac{6}{4} = 1.5 \)
2. \( x_2 = \frac{-3 - 9}{4} = \frac{-12}{4} = -3 \)

Thus, the solutions are:

\[
x_1 = 1.5 \quad \text{and} \quad x_2 = -3
\]

Would you like more details on the process or have any further questions?

Here are some related questions for practice:
1. Solve \( 3x^2 + 7x - 5 = 0 \).
2. Solve \( 4x^2 - 12x + 9 = 0 \).
3. What are the possible types of roots for a quadratic equation?
4. Use the quadratic formula to solve \( x^2 + 5x + 6 = 0 \).
5. How do you derive the quadratic formula?

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

Learn how to solve the quadratic equation 2x² + 3x - 9 using the quadratic formula. This step-by-step solution explains the process, covering key concepts in algebra and quadratic equations. Perfect for students in Grades 8-10.

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