First, let's start by rewriting the quadratic equation \( 3x^2 + 3x + 12 = 10 \) in standard form. We can do this by moving 10 to the left side:
\[ 3x^2 + 3x + 12 - 10 = 0 \]
This simplifies to:
\[ 3x^2 + 3x + 2 = 0 \]
Next, we can use the quadratic formula to find the solutions. The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 3 \), \( b = 3 \), and \( c = 2 \). Now, we will calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 3^2 - 4 \cdot 3 \cdot 2 = 9 - 24 = -15 \]
Since the discriminant is negative, the solutions will involve imaginary numbers. Now, we can substitute the values into the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{-15}}{2 \cdot 3} \]
This simplifies to:
\[ x = \frac{-3 \pm i\sqrt{15}}{6} \]
We can further simplify this expression:
\[ x = \frac{-3}{6} \pm \frac{i\sqrt{15}}{6} \]
This simplifies to:
\[ x = -\frac{1}{2} \pm \frac{i\sqrt{15}}{6} \]
Thus, the solutions of the quadratic equation \( 3x^2 + 3x + 12 = 10 \) in the form \( a \pm bi \) are:
\[ x = -\frac{1}{2} \pm \frac{i\sqrt{15}}{6} \]
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