To solve the equation [tex]\(2x^2 = -10x + 12\)[/tex], we first need to rewrite it in standard form (i.e., setting one side to zero).
### Step-by-Step Solution:
1. Rewrite the equation in standard form:
[tex]\[
2x^2 + 10x - 12 = 0
\][/tex]
2. Factor the quadratic equation:
To factor [tex]\(2x^2 + 10x - 12\)[/tex], we look for two numbers that multiply to [tex]\(2 \times -12 = -24\)[/tex] and add to [tex]\(10\)[/tex]. After some consideration, we find that these numbers are [tex]\(12\)[/tex] and [tex]\(-2\)[/tex].
So we can rewrite the middle term [tex]\(10x\)[/tex] using these numbers:
[tex]\[
2x^2 + 12x - 2x - 12 = 0
\][/tex]
3. Group terms to factor by grouping:
[tex]\[
(2x^2 + 12x) + (-2x - 12) = 0
\][/tex]
4. Factor out the common terms in each group:
[tex]\[
2x(x + 6) - 2(x + 6) = 0
\][/tex]
5. Factor out the common binomial factor:
[tex]\[
(2x - 2)(x + 6) = 0
\][/tex]
6. Solve for [tex]\(x\)[/tex] by setting each factor equal to zero:
[tex]\[
2x - 2 = 0 \quad \text{or} \quad x + 6 = 0
\][/tex]
7. Solve each equation for [tex]\(x\)[/tex]:
[tex]\[
2x - 2 = 0 \implies 2x = 2 \implies x = 1
\][/tex]
[tex]\[
x + 6 = 0 \implies x = -6
\][/tex]
So, the solutions to the equation [tex]\(2x^2 = -10x + 12\)[/tex] are [tex]\(x = 1\)[/tex] and [tex]\(x = -6\)[/tex].
### Verifying Against the Given Options:
- [tex]\(x = 3\)[/tex] : Not a solution
- [tex]\(x = -3\)[/tex] : Not a solution
- [tex]\(x = -2\)[/tex] : Not a solution
- [tex]\(x = 1\)[/tex] : Is a solution
- [tex]\(x = -6\)[/tex] : Is a solution
- [tex]\(x = 6\)[/tex] : Not a solution
Thus, the correct answers are [tex]\( x = 1 \)[/tex] and [tex]\( x = -6 \)[/tex].
