To solve the quadratic equation \(3x^2 + 3x + 12 = 10\), we first rewrite it in standard form:
\[ 3x^2 + 3x + 12 - 10 = 0 \] \[ 3x^2 + 3x + 2 = 0 \]
Next, we can use the quadratic formula to solve for \(x\). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation \(3x^2 + 3x + 2 = 0\), we have:
- \(a = 3\)
- \(b = 3\)
- \(c = 2\)
Now, we can calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 3 \cdot 2 = 9 - 24 = -15 \]
Since the discriminant is negative, this means the solutions will be complex. We will now apply the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{-15}}{2 \cdot 3} \] \[ x = \frac{-3 \pm i\sqrt{15}}{6} \]
We can simplify this expression:
\[ x = \frac{-3}{6} \pm \frac{i\sqrt{15}}{6} \] \[ x = -\frac{1}{2} \pm \frac{i\sqrt{15}}{6} \]
Thus, the solutions in the form \(a \pm bi\) are:
\[ x = -\frac{1}{2} \pm \frac{\sqrt{15}}{6}i \]