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 calculate the discriminant \(b^2 - 4ac\):
\[ b^2 = (-3)^2 = 9 \] \[ 4ac = 4 \cdot 6 \cdot 6 = 144 \] \[ b^2 - 4ac = 9 - 144 = -135 \]
Since the discriminant is negative (\(-135\)), the solutions will be complex numbers.
Now apply the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \cdot 6} \] \[ x = \frac{3 \pm \sqrt{-135}}{12} \]
We can express \(\sqrt{-135}\) in terms of \(i\):
\[ \sqrt{-135} = \sqrt{135}i = \sqrt{9 \cdot 15}i = 3\sqrt{15}i \]
Now substitute this back into the formula:
\[ x = \frac{3 \pm 3\sqrt{15}i}{12} \] \[ x = \frac{3}{12} \pm \frac{3\sqrt{15}i}{12} \] \[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Thus, the solutions of the equation \(6x^2 - 3x + 6 = 0\) are:
\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4} i \]
Among the options provided, this corresponds to:
\[ \text{Start Fraction } 1 \text{ over } 4 \text{ End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i} \]
This is the correct answer.