To find the zeros of the equation [tex]\(2x^2 + 18x - 2 = 6x - 12\)[/tex], follow these steps in a detailed manner until you find the solutions:
1. Rewrite the equation:
Begin by bringing all terms to one side of the equation to set the equation to zero.
[tex]\[
2x^2 + 18x - 2 - 6x + 12 = 0
\][/tex]
2. Simplify the equation:
Combine like terms on the left-hand side.
[tex]\[
2x^2 + (18x - 6x) + (-2 + 12) = 0
\][/tex]
[tex]\[
2x^2 + 12x + 10 = 0
\][/tex]
3. Reduce the equation (if necessary by common factors):
Notice that the coefficients can be divided by 2 to simplify:
[tex]\[
\frac{2}{2}x^2 + \frac{12}{2}x + \frac{10}{2} = 0
\][/tex]
[tex]\[
x^2 + 6x + 5 = 0
\][/tex]
4. Factor the quadratic equation:
Look for two numbers that multiply to 5 (the constant term) and add up to 6 (the coefficient of the linear term). These numbers are 1 and 5.
[tex]\[
(x + 1)(x + 5) = 0
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
Set each factor equal to zero and solve for [tex]\(x\)[/tex].
[tex]\[
x + 1 = 0 \implies x = -1
\][/tex]
[tex]\[
x + 5 = 0 \implies x = -5
\][/tex]
Therefore, the zeros of the equation [tex]\(2x^2 + 18x - 2 = 6x - 12\)[/tex] are:
[tex]\[
x = -1 \quad \text{and} \quad x = -5
\][/tex]
The correct choice from the given options is:
[tex]\[
x = -1 \text{ and } x = -5
\][/tex]
