To solve the equation \((x^2 - x)^2 + 2(x^2 - x) = 8\), we can start by making a substitution. Let:

\[
y = x^2 - x
\]

Now, rewrite the equation in terms of \(y\):

\[
y^2 + 2y - 8 = 0
\]

Next, we can factor this quadratic equation. We want two numbers that multiply to \(-8\) and add up to \(2\). These numbers are \(4\) and \(-2\). Thus, we can factor it as:

\[
(y + 4)(y - 2) = 0
\]

Setting each factor to zero gives us:

1. \(y + 4 = 0 \implies y = -4\)
2. \(y - 2 = 0 \implies y = 2\)

Now, we can substitute back for \(y\):

1. For \(y = -4\):
\[
x^2 - x = -4 \implies x^2 - x + 4 = 0
\]
The discriminant is:
\[
(-1)^2 - 4 \cdot 1 \cdot 4 = 1 - 16 = -15 \quad (\text{no real solutions})
\]

2. For \(y = 2\):
\[
x^2 - x = 2 \implies x^2 - x - 2 = 0
\]
The discriminant is:
\[
(-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \quad (\text{two real solutions})
\]
Solving for \(x\):
\[
x = \frac{1 \pm 3}{2}
\]
This gives:
\[
x = 2 \quad \text{or} \quad x = -1
\]

Thus, the solutions to the original equation are:

\[
x = 2 \quad \text{and} \quad x = -1
\]