Solve the equation: \( |3x^2 - 2x - 1| = 2 \)

Step 1 ：When solving equations involving absolute values, we can rewrite the equation as two separate equations: one where the expression inside the absolute value is equal to the value on the other side of the equation, and one where the expression inside the absolute value is equal to the negative of the value on the other side of the equation.

Step 2 ：So, we have two equations to solve: \(3x^2 - 2x - 1 = 2\) and \(3x^2 - 2x - 1 = -2\)

Step 3 ：Solving the first equation, we subtract 2 from both sides to get \(3x^2 - 2x - 3 = 0\). We can solve this quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), and \(c = -3\). This gives us \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 * 3 * -3}}{2 * 3}\) = \(\frac{2 \pm \sqrt{4 + 36}}{6}\) = \(\frac{2 \pm \sqrt{40}}{6}\) = \(\frac{2 \pm 2\sqrt{10}}{6}\), simplifying to \(x = \frac{1 \pm \sqrt{10}}{3}\)

Step 4 ：Solving the second equation, we add 2 to both sides to get \(3x^2 - 2x + 1 = 0\). Using the quadratic formula again, we get: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4 * 3 * 1}}{2 * 3}\) = \(\frac{2 \pm \sqrt{4 - 12}}{6}\) = \(\frac{2 \pm \sqrt{-8}}{6}\). This equation has no real solutions because the discriminant (the value inside the square root) is negative.