To solve the quadratic equation \( 2x^2 + 4x = -12 \), we first rewrite it in the standard form \( ax^2 + bx + c = 0 \).
Starting with the equation:
\[ 2x^2 + 4x + 12 = 0 \]
Now we can simplify this equation by dividing every term by 2:
\[ x^2 + 2x + 6 = 0 \]
Next, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1 \), \( b = 2 \), and \( c = 6 \). Plugging these into the formula gives:
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \]
Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20 \]
Since the discriminant is negative, we have complex solutions. Now substituting back into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{-20}}{2} \]
To simplify \( \sqrt{-20} \):
\[ \sqrt{-20} = \sqrt{20} i = \sqrt{4 \cdot 5} i = 2\sqrt{5} i \]
Now substituting this back into the equation:
\[ x = \frac{-2 \pm 2\sqrt{5} i}{2} \]
This simplifies to:
\[ x = -1 \pm \sqrt{5} i \]
Thus, the solutions are:
\(-1 \pm i\sqrt{5}\)
The correct choice from the given options is:
−1±i5–√ (negative 1 plus or minus \( i \sqrt{5} \))