To solve the equation \(2(3x + 1)^2 = (3x + 1)(x - 2)\), let's first simplify and solve it step-by-step.

1. **Expand both sides of the equation:**

Left side:
\[
2(3x + 1)^2
\]
First, expand \((3x + 1)^2\):
\[
(3x + 1)^2 = (3x + 1)(3x + 1) = 9x^2 + 6x + 1
\]
Now multiply by 2:
\[
2(9x^2 + 6x + 1) = 18x^2 + 12x + 2
\]

Right side:
\[
(3x + 1)(x - 2)
\]
Expand the product:
\[
(3x + 1)(x - 2) = 3x(x - 2) + 1(x - 2) = 3x^2 - 6x + x - 2 = 3x^2 - 5x - 2
\]

2. **Set the expanded forms equal to each other:**
\[
18x^2 + 12x + 2 = 3x^2 - 5x - 2
\]

3. **Move all terms to one side to set the equation to 0:**
\[
18x^2 + 12x + 2 - 3x^2 + 5x + 2 = 0
\]
Simplify:
\[
15x^2 + 17x + 4 = 0
\]

4. **Solve the quadratic equation \(15x^2 + 17x + 4 = 0\):**

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 15\), \(b = 17\), and \(c = 4\).

Calculate the discriminant:
\[
b^2 - 4ac = 17^2 - 4(15)(4) = 289 - 240 = 49
\]

Now, apply the quadratic formula:
\[
x = \frac{-17 \pm \sqrt{49}}{2 \cdot 15} = \frac{-17 \pm 7}{30}
\]

This gives us two solutions:
\[
x = \frac{-17 + 7}{30} = \frac{-10}{30} = -\frac{1}{3}
\]
\[
x = \frac{-17 - 7}{30} = \frac{-24}{30} = -\frac{4}{5}
\]

So, the solutions to the equation \(2(3x + 1)^2 = (3x + 1)(x - 2)\) are:
\[
x = -\frac{1}{3} \quad \text{and} \quad x = -\frac{4}{5}
\]