To solve the quadratic equation \( 6x^2 - 3x + 6 = 0 \), we can use the quadratic formula given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = -3 \), and \( c = 6 \).
First, 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, we will have complex solutions. Now substitute the values into the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \cdot 6} \] \[ x = \frac{3 \pm \sqrt{-135}}{12} \]
Next, simplify \( \sqrt{-135} \):
\[ \sqrt{-135} = \sqrt{135}i = \sqrt{9 \cdot 15}i = 3\sqrt{15}i \]
Now replace \( \sqrt{-135} \) in the quadratic formula:
\[ x = \frac{3 \pm 3\sqrt{15}i}{12} \]
This can be simplified further:
\[ x = \frac{3}{12} \pm \frac{3\sqrt{15}}{12}i \] \[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Expressing this in the requested form \( a \pm bi \), we get:
\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
Thus, the correct option is:
\[ \frac{1}{4} \pm \frac{\sqrt{15}}{4}i \]
So the final answer is:
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i