To solve for the roots of the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex], we will use the quadratic formula, which is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\( a = 6 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -4 \)[/tex].
### Step-by-Step Solution:
1. Compute the Discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
\text{Discriminant} = 5^2 - 4 \cdot 6 \cdot (-4) = 25 + 96 = 121
\][/tex]
2. Discriminant Square Root:
[tex]\[
\sqrt{121} = 11
\][/tex]
3. Apply the Quadratic Formula:
We have:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute the discriminant and the coefficients:
[tex]\[
x = \frac{-5 \pm 11}{2 \cdot 6} = \frac{-5 \pm 11}{12}
\][/tex]
4. Calculate the Two Roots:
- For the positive root ([tex]\( + \sqrt{\text{Discriminant}} \)[/tex]):
[tex]\[
x_1 = \frac{-5 + 11}{12} = \frac{6}{12} = \frac{1}{2}
\][/tex]
- For the negative root ([tex]\( - \sqrt{\text{Discriminant}} \)[/tex]):
[tex]\[
x_2 = \frac{-5 - 11}{12} = \frac{-16}{12} = -\frac{4}{3} = -1 \frac{1}{3}
\][/tex]
### Conclusion:
The roots of the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex] are:
[tex]\[
x_1 = \frac{1}{2} \quad \text{and} \quad x_2 = -1 \frac{1}{3}
\][/tex]
Thus, the correct option is:
[tex]\[
\boxed{\frac{1}{2} \text{ and } -1 \frac{1}{3}}
\][/tex]
