1. **State the problem:** Solve the equation $$\frac{3}{x+2} = 1 + \frac{2}{x-3}$$ for $x$. 2. **Rewrite the equation:** $$\frac{3}{x+2} = 1 + \frac{2}{x-3}$$ 3. **Bring all terms to one side to set the equation to zero:** $$\frac{3}{x+2} - 1 - \frac{2}{x-3} = 0$$ 4. **Find a common denominator:** The denominators are $(x+2)$ and $(x-3)$, so the common denominator is $(x+2)(x-3)$. 5. **Rewrite each term with the common denominator:** $$\frac{3(x-3)}{(x+2)(x-3)} - \frac{(x+2)(x-3)}{(x+2)(x-3)} - \frac{2(x+2)}{(x+2)(x-3)} = 0$$ 6. **Combine the fractions:** $$\frac{3(x-3) - (x+2)(x-3) - 2(x+2)}{(x+2)(x-3)} = 0$$ 7. **Since the denominator cannot be zero, set the numerator equal to zero:** $$3(x-3) - (x+2)(x-3) - 2(x+2) = 0$$ 8. **Expand each term:** $$3x - 9 - (x^2 - 3x + 2x - 6) - 2x - 4 = 0$$ 9. **Simplify inside the parentheses:** $$3x - 9 - (x^2 - x - 6) - 2x - 4 = 0$$ 10. **Distribute the minus sign:** $$3x - 9 - x^2 + x + 6 - 2x - 4 = 0$$ 11. **Combine like terms:** $$-x^2 + (3x + x - 2x) + (-9 + 6 - 4) = 0$$ $$-x^2 + 2x - 7 = 0$$ 12. **Multiply both sides by $-1$ to simplify:** $$\cancel{-}x^2 + 2x - 7 = 0 \Rightarrow x^2 - 2x + 7 = 0$$ 13. **Use the quadratic formula to solve:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-2$, $c=7$. 14. **Calculate the discriminant:** $$\Delta = (-2)^2 - 4(1)(7) = 4 - 28 = -24$$ 15. **Since the discriminant is negative, the solutions are complex:** $$x = \frac{2 \pm \sqrt{-24}}{2} = \frac{2 \pm 2i\sqrt{6}}{2} = 1 \pm i\sqrt{6}$$ **Final answer:** $$x = 1 + i\sqrt{6} \quad \text{or} \quad x = 1 - i\sqrt{6}$$