1. **State the problem:** Solve the equation $$(4x+1)(x+3)-3(x-2)=(2x-3)^2$$ for $x$. 2. **Expand both sides:** - Left side: $$(4x+1)(x+3) = 4x^2 + 12x + x + 3 = 4x^2 + 13x + 3$$ - Also subtract $3(x-2) = 3x - 6$, so left side becomes $$4x^2 + 13x + 3 - 3x + 6 = 4x^2 + 10x + 9$$ - Right side: $$(2x-3)^2 = (2x)^2 - 2 \cdot 2x \cdot 3 + 3^2 = 4x^2 - 12x + 9$$ 3. **Set the equation:** $$4x^2 + 10x + 9 = 4x^2 - 12x + 9$$ 4. **Simplify by subtracting $4x^2$ and $9$ from both sides:** $$10x = -12x$$ 5. **Combine like terms:** $$10x + 12x = 0 \implies 22x = 0$$ 6. **Solve for $x$:** $$x = 0$$ **Final answer:** $$x = 0$$