We can solve for x using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a, b, and c are the coefficients of the quadratic equation.

In this case, a = 3, b = 6, and c = -2. Substituting these values into the quadratic formula gives:

x = \(\frac{-6 \pm \sqrt{(6)^2 - 4*3*(-2)}}{2*3}\)
= \(\frac{-6 \pm \sqrt{36 + 24}}{6}\)
= \(\frac{-6 \pm \sqrt{60}}{6}\)
= \(\frac{-6 \pm 2\sqrt{15}}{6}\)
= -1 \(\pm\) \(\frac{\sqrt{15}}{3}\)

Hence, the solutions to the equation \(3x^2 + 6x - 2 = 0\) are x = -1 + \(\frac{\sqrt{15}}{3}\) and x = -1 - \(\frac{\sqrt{15}}{3}\).