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Let's solve these equations step by step:

### a. \( x^3 = -i \)

To solve this, we'll find the cube roots of the complex number \(-i\).

1. Express \(-i\) in polar form: \(-i = 1 \cdot e^{-i\frac{\pi}{2}}\).
2. Find the cube roots using the formula \(x_k = r^{1/3} \cdot e^{i(\theta + 2k\pi)/3}\), where \(k = 0, 1, 2\).

   Thus, we get three roots:
   \[
   x_0 = e^{-i\frac{\pi}{6}}, \quad x_1 = e^{i\frac{5\pi}{6}}, \quad x_2 = e^{i\frac{13\pi}{6}}.
   \]
   These can be converted back to rectangular form.

### b. \( 2x^3 + 14x^2 + 32x + 20 = 0 \)

Use the Rational Root Theorem to find possible rational roots, and then perform synthetic division or factorization to simplify the polynomial.

- Checking possible roots like \(x = -2\), we find \(x + 2\) is a factor.
- Factorize to get the other roots.

### c. \( x^2 = -3 \)

Solve for \(x\):
\[
x = \pm \sqrt{-3} = \pm i\sqrt{3}.
\]

### d. \( x^2 - 5x - 24 = 0 \)

Use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \(a = 1\), \(b = -5\), and \(c = -24\). This gives the solutions.

### e. \( 3x^2 + 22 = 0 \)

Rewrite it as:
\[
x^2 = -\frac{22}{3},
\]
giving the roots:
\[
x = \pm i\sqrt{\frac{22}{3}}.
\]

Would you like further details on any of these solutions or a different approach?

### Relative Questions:
1. How do we convert a

Resuelve las ecuaciones:
 a. x^3 = -i 
 b. 2x^3 + 14x^2 + 32x + 20 = 0 
 c. x^2 = -3 
 d. x^2 - 5x - 24 = 0 
 e. 3x^2 + 22 = 0

Step-by-step solutions to equations involving complex and imaginary numbers: x^3 = -i, quadratic polynomials, and more. Learn how to apply key algebraic and trigonometric concepts, suitable for Grades 9-12.

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