To solve the equation \((x^2 + x)^2 + 4(x^2 + x) - 12 = 0\), we can simplify it by letting \( y = x^2 + x \). This substitution will make the equation a quadratic in terms of \( y \).

So, the equation becomes:
\[ y^2 + 4y - 12 = 0 \]

Next, we solve the quadratic equation for \( y \). We use the quadratic formula \( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -12 \):
\[ y = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 1 \cdot (-12)}}}}{2 \cdot 1} \]
\[ y = \frac{{-4 \pm \sqrt}}{2} \]
\[ y = \frac{{-4 \pm \sqrt{64}}}{2} \]
\[ y = \frac{{-4 \pm 8}}{2} \]

This gives us two solutions for \( y \):
\[ y = \frac{{-4 + 8}}{2} = 2 \]
\[ y = \frac{{-4 - 8}}{2} = -6 \]

Now, we revert our substitution \( y = x^2 + x \), yielding two cases:
1. \( x^2 + x = 2 \)
2. \( x^2 + x = -6 \)

### Case 1: \( x^2 + x = 2 \)

This simplifies to:
\[ x^2 + x - 2 = 0 \]

We solve this quadratic equation using the quadratic formula:
\[ x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot (-2)}}}}{2 \cdot 1} \]
\[ x = \frac{{-1 \pm \sqrt}}{2} \]
\[ x = \frac{{-1 \pm \sqrt{9}}}{2} \]
\[ x = \frac{{-1 \pm 3}}{2} \]

This gives us:
\[ x = \frac{{-1 + 3}}{2} = 1 \]
\[ x = \frac{{-1 - 3}}{2} = -2 \]

### Case 2: \( x^2 + x = -6 \)

This simplifies to:
\[ x^2 + x + 6 = 0 \]

We solve this quadratic equation using the quadratic formula:
\[ x = \frac{{-1 \pm \sqrt{{1^2 - 4 \cdot 1 \cdot 6}}}}{2 \cdot 1} \]
\[ x = \frac{{-1 \pm \sqrt}}{2} \]
\[ x = \frac{{-1 \pm \sqrt{-23}}}{2} \]

Since the discriminant (\(\sqrt{-23}\)) is negative, there are no real solutions for \( x \) in this case.

### Summary

The real solutions to the original equation are:
\[ x = 1 \]
\[ x = -2 \]