To solve the equation \( x^3 + x^2 + 4 = 0 \), we can start by seeing if there are any rational roots using the Rational Root Theorem but based on inspection and the nature of the polynomial, it seems that this polynomial may not have rational roots.

Let's check for possible roots by evaluating the function:

1. Compute \( f(-2) \):
\[
f(-2) = (-2)^3 + (-2)^2 + 4 = -8 + 4 + 4 = 0
\]
Thus, \( x = -2 \) is a root.

Now that we have found one root, we can factor the polynomial by using synthetic division to divide \( x^3 + x^2 + 4 \) by \( x + 2 \):

### Synthetic Division:

```
-2 | 1 1 0 4
| -2 2 -4
-------------------
1 -1 2 0
```

The quotient is \( x^2 - x + 2 \), so we can write the original polynomial as:
\[
x^3 + x^2 + 4 = (x + 2)(x^2 - x + 2)
\]

Next, we need to find the roots of \( x^2 - x + 2 = 0 \). We will use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1, b = -1, c = 2 \).

Calculating the discriminant:
\[
b^2 - 4ac = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7
\]

Since the discriminant is negative, \( x^2 - x + 2 \) has no real roots and has complex roots. We can calculate these complex roots:
\[
x = \frac{1 \pm \sqrt{-7}}{2} = \frac{1 \pm i\sqrt{7}}{2}
\]

### Final Roots:
1. The real root is \( x = -2 \).
2. The complex roots are \( x = \frac{1 + i\sqrt{7}}{2} \) and \( x = \frac{1 - i\sqrt{7}}{2} \).

In conclusion, the solutions to the equation \( x^3 + x^2 + 4 = 0 \) are:
\[
x = -2, \quad x = \frac{1 + i\sqrt{7}}{2}, \quad x = \frac{1 - i\sqrt{7}}{2}
\]