To solve the quadratic equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex], we'll need to follow several steps:
1. Rewrite the equation in standard form:
The given equation is already in standard form: [tex]\(2x^2 + 10x - 12 = 0\)[/tex].
2. Identify the coefficients:
Here, [tex]\(a = 2\)[/tex], [tex]\(b = 10\)[/tex], and [tex]\(c = -12\)[/tex].
3. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] is given by [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]
4. Determine the solutions using the quadratic formula:
The quadratic formula is [tex]\( x = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex].
Since [tex]\(\Delta = 196\)[/tex],
[tex]\[
x = \frac{-10 \pm \sqrt{196}}{2 \cdot 2}
\][/tex]
[tex]\[
x = \frac{-10 \pm 14}{4}
\][/tex]
5. Calculate the two solutions:
[tex]\[
x_1 = \frac{-10 + 14}{4} = \frac{4}{4} = 1
\][/tex]
[tex]\[
x_2 = \frac{-10 - 14}{4} = \frac{-24}{4} = -6
\][/tex]
Therefore, the solutions to the equation [tex]\(2x^2 + 10x - 12 = 0\)[/tex] are [tex]\(x = 1\)[/tex] and [tex]\(x = -6\)[/tex].
Given the choices:
- [tex]\(x = -2\)[/tex]
- [tex]\(x = -6\)[/tex]
- [tex]\(x = -3\)[/tex]
- [tex]\(x = 1\)[/tex]
- [tex]\(x = 3\)[/tex]
- [tex]\(x = 6\)[/tex]
The correct solutions are [tex]\(x = 1\)[/tex] and [tex]\(x = -6\)[/tex]. Therefore, you should select:
- [tex]\(x = -6\)[/tex]
- [tex]\(x = 1\)[/tex].
