To find the solutions to the quadratic equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex], we can use the quadratic formula, which states that for any quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], the solutions for [tex]\(x\)[/tex] can be found using:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, our equation is [tex]\(6x^2 + 5x - 4 = 0\)[/tex]. We can identify the coefficients as follows:
- [tex]\(a = 6\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = -4\)[/tex]
Plugging these values into the quadratic formula, we get:
[tex]\[
x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 6 \cdot (-4)}}{2 \cdot 6}
\][/tex]
Let's calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]) first:
[tex]\[
b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot (-4) = 25 + 96 = 121
\][/tex]
So, we have:
[tex]\[
x = \frac{-5 \pm \sqrt{121}}{12} = \frac{-5 \pm 11}{12}
\][/tex]
This results in two solutions:
1. For the positive root:
[tex]\[
x = \frac{-5 + 11}{12} = \frac{6}{12} = \frac{1}{2}
\][/tex]
2. For the negative root:
[tex]\[
x = \frac{-5 - 11}{12} = \frac{-16}{12} = -\frac{4}{3}
\][/tex]
Therefore, the solutions to the equation [tex]\(6x^2 + 5x - 4 = 0\)[/tex] are [tex]\(x = \frac{1}{2}\)[/tex] and [tex]\(x = -\frac{4}{3}\)[/tex].
From the given options:
- [tex]\(x = 4\)[/tex]
- [tex]\(x = -\frac{4}{3}\)[/tex]
- [tex]\(x = \frac{1}{2}\)[/tex]
- [tex]\(x = -2\)[/tex]
- [tex]\(x = -\frac{1}{3}\)[/tex]
- [tex]\(x = \frac{1}{3}\)[/tex]
We see that the correct solutions that match the options are:
- [tex]\(x = \frac{1}{2}\)[/tex]
- [tex]\(x = -\frac{4}{3}\)[/tex]
Thus, the correct answers are:
B. [tex]\(x = -\frac{4}{3}\)[/tex]
C. [tex]\(x = \frac{1}{2}\)[/tex]
