To solve the quadratic equation \( 3x^2 + 3x + 12 = 10 \), we first rearrange it into standard form:
\[ 3x^2 + 3x + 12 - 10 = 0 \]
This simplifies to:
\[ 3x^2 + 3x + 2 = 0 \]
Next, we will use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, we identify \( a = 3 \), \( b = 3 \), and \( c = 2 \). We will first calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 3^2 - 4(3)(2) = 9 - 24 = -15 \]
Since the discriminant is negative, the solutions will be complex. We can now apply the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{-15}}{2(3)} \]
This simplifies to:
\[ x = \frac{-3 \pm \sqrt{15}i}{6} \]
We can further simplify this:
\[ x = \frac{-3}{6} \pm \frac{\sqrt{15}i}{6} \]
Which gives:
\[ x = -\frac{1}{2} \pm \frac{\sqrt{15}}{6}i \]
Thus, the solutions of the quadratic equation \( 3x^2 + 3x + 12 = 10 \) are:
\[ x = -\frac{1}{2} \pm \frac{\sqrt{15}}{6}i \]
This is the final answer in the form \( a \pm bi \).