1. **State the problem:** Solve the quadratic equation $$3x^2 + 3x - 6 = 0$$. 2. **Formula used:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. **Identify coefficients:** Here, $a = 3$, $b = 3$, and $c = -6$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 3 \times (-6) = 9 + 72 = 81$$. 5. **Apply the quadratic formula:** $$x = \frac{-3 \pm \sqrt{81}}{2 \times 3} = \frac{-3 \pm 9}{6}$$. 6. **Find the two solutions:** - For the plus sign: $$x = \frac{-3 + 9}{6} = \frac{6}{6} = 1$$. - For the minus sign: $$x = \frac{-3 - 9}{6} = \frac{-12}{6} = -2$$. 7. **Final answer:** The solutions to the equation are $$x = 1$$ and $$x = -2$$.