Solve the quadratic equation 2x2+4x=−12

To solve the quadratic equation \(2x^2 + 4x = -12\), we first rearrange it into standard form:

\[ 2x^2 + 4x + 12 = 0 \]

Next, we can simplify the equation by dividing all terms by 2:

\[ x^2 + 2x + 6 = 0 \]

Now we will use the quadratic formula to find the solutions:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our equation, \(a = 1\), \(b = 2\), and \(c = 6\). Plugging these values into the quadratic formula gives:

\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20 \]

Since the discriminant is negative, we will have complex solutions.

Now, substituting back into the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{-20}}{2 \cdot 1} \]

We can rewrite \(\sqrt{-20}\) as \(\sqrt{20}i\):

\[ \sqrt{-20} = \sqrt{20}i = \sqrt{4 \cdot 5}i = 2\sqrt{5}i \]

So, substituting this back into the formula:

\[ x = \frac{-2 \pm 2\sqrt{5}i}{2} \]

Simplifying this yields:

\[ x = -1 \pm \sqrt{5}i \]

Thus, the solutions in the form \(a \pm bi\) are:

\[ -1 \pm i\sqrt{5} \]

So the correct answer from the provided options is:

−1±i5–√ (negative 1 plus or minus i Start Root 5 End Root)