To solve the quadratic equation $x^2 + 4x + 13 = 0$ over the complex numbers, we use the quadratic formula:

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

Here, $a = 1$, $b = 4$, and $c = 13$. Substituting these values:

$x = \frac{-4 \pm \sqrt{4^2 - 4(1)(13)}}{2(1)}$

$x = \frac{-4 \pm \sqrt{16 - 52}}{2}$

$x = \frac{-4 \pm \sqrt{-36}}{2}$

Since $\sqrt{-36} = 6i$ (where $i$ is the imaginary unit), we have:

$x = \frac{-4 \pm 6i}{2}$

Simplify the expression:

$x = -2 \pm 3i$

Solutions:

The two solutions are:

1. $x = -2 + 3i$
2. $x = -2 - 3i$

Would you like a detailed explanation of any step?

Related Questions:

1. What are the properties of the discriminant in quadratic equations with complex roots?
2. How does the imaginary unit $i$ simplify calculations involving square roots of negative numbers?
3. Can you explain why this quadratic equation does not have real solutions?
4. How does completing the square compare to using the quadratic formula for this problem?
5. What practical applications involve complex solutions to quadratic equations?

Tip:

Always check the discriminant $b^2 - 4ac$ to quickly determine whether the solutions are real, complex, or repeated roots.