To solve the quadratic equation \(2x^2 + 4x = -12\), we start by rearranging it into standard form:
\[ 2x^2 + 4x + 12 = 0. \]
Next, we simplify this equation by dividing all terms by 2:
\[ x^2 + 2x + 6 = 0. \]
Now, we will use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \(a = 1\), \(b = 2\), and \(c = 6\).
First, we calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20. \]
Since the discriminant is negative, this indicates that the solutions will be complex numbers. We can now plug in the values into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{-20}}{2 \cdot 1}. \]
We can simplify this further:
\[ x = \frac{-2 \pm \sqrt{20}i}{2}. \]
Since \(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\), we have:
\[ x = \frac{-2 \pm 2\sqrt{5}i}{2} = -1 \pm \sqrt{5}i. \]
Now, we can express the solutions as:
\[ x = -1 \pm i\sqrt{5}. \]
Thus, the correct answer is:
\(-1 \pm i\sqrt{5}\).