1. **State the problem:** Find the solution set of the equation $$x + \frac{4}{x+3} = 2$$. 2. **Rewrite the equation:** To solve for $x$, first eliminate the fraction by multiplying both sides by $x+3$ (noting $x \neq -3$ to avoid division by zero): $$ (x+3) \left(x + \frac{4}{x+3}\right) = 2(x+3) $$ 3. **Distribute and simplify:** $$ (x+3) x + (x+3) \frac{4}{x+3} = 2x + 6 $$ $$ x^2 + 3x + 4 = 2x + 6 $$ 4. **Bring all terms to one side:** $$ x^2 + 3x + 4 - 2x - 6 = 0 $$ $$ x^2 + x - 2 = 0 $$ 5. **Factor the quadratic:** $$ x^2 + x - 2 = (x + 2)(x - 1) = 0 $$ 6. **Solve for $x$:** $$ x + 2 = 0 \Rightarrow x = -2 $$ $$ x - 1 = 0 \Rightarrow x = 1 $$ 7. **Check for restrictions:** Since $x \neq -3$ (denominator restriction), both $x = -2$ and $x = 1$ are valid. **Final answer:** The solution set is $$\{-2, 1\}$$ which corresponds to choice A.