1. **State the problem:** Solve the equation $$(3x+1)^2 - (2x-5)^2 = 504$$ for $x$. 2. **Recall the difference of squares formula:** $$a^2 - b^2 = (a-b)(a+b)$$ This formula helps simplify expressions like the one given. 3. **Apply the formula:** Let $a = 3x+1$ and $b = 2x-5$. Then, $$(3x+1)^2 - (2x-5)^2 = ((3x+1) - (2x-5))((3x+1) + (2x-5))$$ 4. **Simplify each factor:** $$((3x+1) - (2x-5)) = 3x + 1 - 2x + 5 = x + 6$$ $$((3x+1) + (2x-5)) = 3x + 1 + 2x - 5 = 5x - 4$$ 5. **Rewrite the equation:** $$(x + 6)(5x - 4) = 504$$ 6. **Expand the left side:** $$5x^2 - 4x + 30x - 24 = 504$$ $$5x^2 + 26x - 24 = 504$$ 7. **Bring all terms to one side:** $$5x^2 + 26x - 24 - 504 = 0$$ $$5x^2 + 26x - 528 = 0$$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=26$, and $c=-528$. Calculate the discriminant: $$\Delta = 26^2 - 4 \times 5 \times (-528) = 676 + 10560 = 11236$$ Calculate the square root: $$\sqrt{11236} = 106$$ 9. **Find the roots:** $$x = \frac{-26 \pm 106}{2 \times 5} = \frac{-26 \pm 106}{10}$$ Two solutions: $$x_1 = \frac{-26 + 106}{10} = \frac{80}{10} = 8$$ $$x_2 = \frac{-26 - 106}{10} = \frac{-132}{10} = -13.2$$ 10. **Final answer:** $$x = 8 \quad \text{or} \quad x = -13.2$$