1. **State the problem:** Solve the equation $$(2x+3)^2 = 12x$$ for $x$. 2. **Use the formula:** Recall that $$(a+b)^2 = a^2 + 2ab + b^2$$. 3. **Expand the left side:** $$ (2x+3)^2 = (2x)^2 + 2 \cdot 2x \cdot 3 + 3^2 = 4x^2 + 12x + 9 $$ 4. **Rewrite the equation:** $$ 4x^2 + 12x + 9 = 12x $$ 5. **Bring all terms to one side:** $$ 4x^2 + 12x + 9 - 12x = 0 $$ $$ 4x^2 + \cancel{12x} + 9 - \cancel{12x} = 0 $$ $$ 4x^2 + 9 = 0 $$ 6. **Solve for $x^2$:** $$ 4x^2 = -9 $$ $$ x^2 = \frac{-9}{4} $$ 7. **Interpret the result:** Since $x^2$ equals a negative number, there are no real solutions. 8. **Write the complex solutions:** $$ x = \pm \sqrt{\frac{-9}{4}} = \pm \frac{3i}{2} $$ **Final answer:** $$ x = \frac{3i}{2} \quad \text{or} \quad x = -\frac{3i}{2} $$