To solve the equation \((x^2 + x + 1)(6 + 2x) = 0\), we can set each factor equal to zero.

1. **First factor**: \(x^2 + x + 1 = 0\)

To find the roots of this quadratic equation, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = 1\), and \(c = 1\).

Calculating the discriminant:
\[
b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3
\]
Since the discriminant is negative, the solutions will be complex numbers:
\[
x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
\]
Thus, the roots from the first factor are:
\[
x = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2}
\]

2. **Second factor**: \(6 + 2x = 0\)

Solving for \(x\):
\[
2x = -6 \implies x = -3
\]

Combining the solutions from both factors, the complete set of solutions to the equation \((x^2 + x + 1)(6 + 2x) = 0\) is:
\[
x = -3, \quad x = \frac{-1 + i\sqrt{3}}{2}, \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2}
\]