To solve for [tex]\( x \)[/tex] in the equation [tex]\( 6x^2 = 2x - 1 \)[/tex], we need to rewrite the equation in the standard quadratic form, which is [tex]\( ax^2 + bx + c = 0 \)[/tex].
Starting with:
[tex]\[ 6x^2 = 2x - 1 \][/tex]
First, bring all terms to one side:
[tex]\[ 6x^2 - 2x + 1 = 0 \][/tex]
The quadratic equation is:
[tex]\[ 6x^2 - 2x + 1 = 0 \][/tex]
Now, we identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ a = 6 \][/tex]
[tex]\[ b = -2 \][/tex]
[tex]\[ c = 1 \][/tex]
Next, we use the quadratic formula to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 - 24}}{12} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{-20}}{12} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{20}i}{12} \][/tex]
[tex]\[ x = \frac{2 \pm 2\sqrt{5}i}{12} \][/tex]
[tex]\[ x = \frac{2 + 2\sqrt{5}i}{12}, \quad \frac{2 - 2\sqrt{5}i}{12} \][/tex]
[tex]\[ x = \frac{1 + \sqrt{5}i}{6}, \quad \frac{1 - \sqrt{5}i}{6} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{1 \pm i\sqrt{5}}{6} \][/tex]
Comparing the solutions with the options provided:
- [tex]\( x = \frac{1 \pm i\sqrt{5}}{6} \)[/tex] is the correct solution.
Therefore, the answer is [tex]\( \boxed{1} \)[/tex].
