To solve the quadratic equation \(2a^2 + 13a + 15 = 0\), we can use the quadratic formula:

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

Here, the coefficients are:
- \(a = 2\)
- \(b = 13\)
- \(c = 15\)

Let's plug these values into the quadratic formula:

\[
a = \frac{-13 \pm \sqrt{13^2 - 4(2)(15)}}{2(2)}
\]

\[
a = \frac{-13 \pm \sqrt{169 - 120}}{4}
\]

\[
a = \frac{-13 \pm \sqrt{49}}{4}
\]

\[
a = \frac{-13 \pm 7}{4}
\]

This gives us two possible solutions:

1. \(a_1 = \frac{-13 + 7}{4} = \frac{-6}{4} = -\frac{3}{2}\)
2. \(a_2 = \frac{-13 - 7}{4} = \frac{-20}{4} = -5\)

So the solutions are:

\[
a = -\frac{3}{2} \quad \text{or} \quad a = -5
\]

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. How do you derive the quadratic formula?
2. What is the discriminant, and how does it determine the nature of the roots?
3. Can you solve quadratic equations by factoring instead of using the quadratic formula?
4. What are the applications of quadratic equations in real-world problems?
5. How do you solve a quadratic equation with complex roots?

**Tip:** When solving quadratic equations, always check the discriminant (\(b^2 - 4ac\)) first to understand the nature of the roots.

Discover how to solve the quadratic equation 2a^2 + 13a + 15 = 0 using the quadratic formula. This step-by-step solution covers essential algebraic concepts and the application of the quadratic formula, ideal for students in Grades 9-10.

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