1. **State the problem:** Solve the equation $$(2x + 5)^2 = (2x + 3)(2x - 1)$$ for $x$. 2. **Use the formula and expand both sides:** Expand the left side using the formula $$(a+b)^2 = a^2 + 2ab + b^2$$: $$ (2x + 5)^2 = (2x)^2 + 2 \cdot 2x \cdot 5 + 5^2 = 4x^2 + 20x + 25 $$ Expand the right side using distributive property: $$ (2x + 3)(2x - 1) = 2x \cdot 2x + 2x \cdot (-1) + 3 \cdot 2x + 3 \cdot (-1) = 4x^2 - 2x + 6x - 3 = 4x^2 + 4x - 3 $$ 3. **Set the equation:** $$ 4x^2 + 20x + 25 = 4x^2 + 4x - 3 $$ 4. **Subtract $4x^2$ from both sides to simplify:** $$ \cancel{4x^2} + 20x + 25 = \cancel{4x^2} + 4x - 3 $$ which simplifies to $$ 20x + 25 = 4x - 3 $$ 5. **Bring all terms involving $x$ to one side and constants to the other:** $$ 20x - 4x = -3 - 25 $$ $$ 16x = -28 $$ 6. **Solve for $x$ by dividing both sides by 16:** $$ x = \frac{-28}{16} $$ Simplify the fraction by dividing numerator and denominator by 4: $$ x = \frac{\cancel{-28}^{\div 4}}{\cancel{16}^{\div 4}} = \frac{-7}{4} $$ **Final answer:** $$ \boxed{x = -\frac{7}{4}} $$