Given the quadratic equation:
[tex]\[
3x^2 + 2x = 6
\][/tex]
First, we need to express this equation in standard quadratic form:
[tex]\[
3x^2 + 2x - 6 = 0
\][/tex]
Here, the coefficients are [tex]\(a = 3\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = -6\)[/tex]. We will solve for [tex]\(x\)[/tex] using the quadratic formula:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
1. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot (-6) = 4 + 72 = 76
\][/tex]
2. Take the square root of the discriminant:
[tex]\[
\sqrt{\Delta} = \sqrt{76} \approx 8.72
\][/tex]
3. Apply the quadratic formula to find the roots:
[tex]\[
x_1 = \frac{{-b + \sqrt{\Delta}}}{2a} = \frac{{-2 + 8.72}}{6} = \frac{6.72}{6} \approx 1.12
\][/tex]
[tex]\[
x_2 = \frac{{-b - \sqrt{\Delta}}}{2a} = \frac{{-2 - 8.72}}{6} = \frac{-10.72}{6} \approx -1.79
\][/tex]
Thus, the solutions to the equation [tex]\(3x^2 + 2x - 6 = 0\)[/tex], rounded to two decimal places, are:
[tex]\[
x = 1.12 \quad \text{and} \quad x = -1.79
\][/tex]
From the given options, the correct one is:
[tex]\[
x = 1.12 \quad \text{and} \quad x = -1.79
\][/tex]
