To solve the quadratic equation \(6x^2 - 3x + 6 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 6\), \(b = -3\), and \(c = 6\).
First, we need to calculate the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4(6)(6) = 9 - 144 = -135 \]
Since the discriminant is negative, the solutions will be complex. Proceeding with the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{-135}}{2(6)} = \frac{3 \pm \sqrt{135}i}{12} \]
Next, we simplify \(\sqrt{135}\):
\[ \sqrt{135} = \sqrt{9 \cdot 15} = 3\sqrt{15} \]
Substituting this back into the equation:
\[ x = \frac{3 \pm 3\sqrt{15}i}{12} \]
We can simplify this:
\[ x = \frac{3}{12} \pm \frac{3\sqrt{15}}{12}i = \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Thus, the solutions are:
\[ \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Now, let's match this with the provided options:
The correct option is:
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i