To solve the quadratic equation \(6x^2 + 13x + 6 = 0\), we can use the quadratic formula:

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

where:
- \(a = 6\)
- \(b = 13\)
- \(c = 6\)

Let's substitute these values into the formula.

\[
x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 6 \cdot 6}}{2 \cdot 6}
\]

Now, let's calculate the discriminant:

\[
b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 6 = 169 - 144 = 25
\]

Substitute the discriminant back into the formula:

\[
x = \frac{-13 \pm \sqrt{25}}{12}
\]

Since \(\sqrt{25} = 5\), we get:

\[
x = \frac{-13 \pm 5}{12}
\]

This gives two solutions:

\[
x_1 = \frac{-13 + 5}{12} = \frac{-8}{12} = -\frac{2}{3}
\]

\[
x_2 = \frac{-13 - 5}{12} = \frac{-18}{12} = -\frac{3}{2}
\]

### Solutions:

\[
x = -\frac{2}{3} \quad \text{and} \quad x = -\frac{3}{2}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How do you complete the square for a quadratic equation?
2. What are the steps to factor quadratic equations manually?
3. How is the quadratic formula derived from a standard quadratic equation?
4. What does the discriminant tell you about the nature of the roots?
5. Can a quadratic equation have complex solutions?

**Tip:** The discriminant (\(b^2 - 4ac\)) in the quadratic formula indicates the nature of the roots. If it's positive, there are two real roots; if it's zero, there is one real root; if it's negative, there are two complex roots.

Discover how to solve the quadratic equation 6x² + 13x + 6 using the quadratic formula. This detailed step-by-step guide explores key concepts such as the discriminant and provides a solution suitable for Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation