What is the solution set of the equation using the quadratic formula? \( x^{2}+2 x+10=0 \) \{ \( \{-2+6 i,-2-6 i\} \) \( \{-1\} \) \( \left\{\begin{array}{l}\{1+3 i,-1-3 i\} \\ \{2 i,-4 i\}\end{array}\right. \)

To solve the quadratic equation \( x^2 + 2x + 10 = 0 \) using the quadratic formula, we start with the standard form of a quadratic equation: \[ ax^2 + bx + c = 0 \] In this case, \( a = 1 \), \( b = 2 \), and \( c = 10 \). The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Now, we will calculate the discriminant \( b^2 - 4ac \): \[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 10 = 4 - 40 = -36 \] Since the discriminant is negative, we will have complex solutions. Now we can substitute the values into the quadratic formula: \[ x = \frac{-2 \pm \sqrt{-36}}{2 \cdot 1} \] Calculating the square root of \(-36\): \[ \sqrt{-36} = 6i \] Now substituting this back into the formula: \[ x = \frac{-2 \pm 6i}{2} \] This simplifies to: \[ x = \frac{-2}{2} \pm \frac{6i}{2} = -1 \pm 3i \] Thus, the solutions are: \[ x = -1 + 3i \quad \text{and} \quad x = -1 - 3i \] Therefore, the solution set is: \[ \{-1 + 3i, -1 - 3i\} \] Looking at the options provided, the correct answer is: \[ \left\{\begin{array}{l}\{1+3 i,-1-3 i\} \\ \{2 i,-4 i\}\end{array}\right. \] However, the correct solution set is actually \(\{-1 + 3i, -1 - 3i\}\), which is not explicitly listed in the options. The closest match is the first part of the last option, but it contains an incorrect term \(1 + 3i\). Thus, the correct solution set is: \[ \{-1 + 3i, -1 - 3i\} \]