To solve the quadratic equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex], we can use the quadratic formula, which is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, the coefficients are:
[tex]\[
a = 2, \quad b = 10, \quad c = -12
\][/tex]
First, we calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
[tex]\[
\Delta = 10^2 - 4 \cdot 2 \cdot (-12)
\][/tex]
[tex]\[
\Delta = 100 + 96
\][/tex]
[tex]\[
\Delta = 196
\][/tex]
Since the discriminant [tex]\(\Delta\)[/tex] is positive, there are two distinct real solutions. We now use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a}
\][/tex]
[tex]\[
x = \frac{-10 \pm \sqrt{196}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{-10 \pm 14}{4}
\][/tex]
This gives us two solutions:
1. [tex]\[
x_1 = \frac{-10 + 14}{4} = \frac{4}{4} = 1
\][/tex]
2. [tex]\[
x_2 = \frac{-10 - 14}{4} = \frac{-24}{4} = -6
\][/tex]
Thus, the solutions to the equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = -6
\][/tex]
Therefore, out of the provided options, the correct answers are:
- [tex]\(x = -6\)[/tex]
- [tex]\(x = 1\)[/tex]
