To solve the quadratic equation \( 6x^2 - 3x + 6 = 0 \), we will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = -3 \), and \( c = 6 \).
- First, we calculate \( b^2 - 4ac \):
\[ b^2 = (-3)^2 = 9 \] \[ 4ac = 4 \times 6 \times 6 = 144 \] \[ b^2 - 4ac = 9 - 144 = -135 \]
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Since the determinant \( b^2 - 4ac \) is negative, there will be complex solutions.
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Now we substitute \( a \), \( b \), and the value of the determinant into the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{-135}}{2 \times 6} \] \[ x = \frac{3 \pm \sqrt{-135}}{12} \]
- We can simplify \( \sqrt{-135} \):
\[ \sqrt{-135} = \sqrt{135} \cdot i = \sqrt{9 \cdot 15} \cdot i = 3\sqrt{15} \cdot i \]
- Substitute this back into the equation:
\[ x = \frac{3 \pm 3\sqrt{15} \cdot i}{12} \]
- We can simplify \( \frac{3}{12} \) and \( \frac{3\sqrt{15}}{12} \):
\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4} \cdot i \]
Thus, the solutions can be expressed in the form \( a \pm bi \) as:
\[ x = \frac{1}{4} \pm \frac{\sqrt{15}}{4} i \]
Therefore, the correct response from your options is:
Start Fraction 1 over 4 End Fraction plus or minus Start Fraction Start Root 15 End Root over 4 End Fraction i