1. **Problem Statement:** Solve a hard radical equation such as $$\sqrt{2x+3} + \sqrt{x-1} = 5$$. 2. **Formula and Rules:** To solve radical equations, isolate one radical on one side, then square both sides to eliminate the radical. Repeat if necessary. Always check for extraneous solutions by substituting back. 3. **Step 1: Isolate one radical:** $$\sqrt{2x+3} = 5 - \sqrt{x-1}$$ 4. **Step 2: Square both sides:** $$\left(\sqrt{2x+3}\right)^2 = \left(5 - \sqrt{x-1}\right)^2$$ $$2x+3 = 25 - 10\sqrt{x-1} + (x-1)$$ 5. **Step 3: Simplify:** $$2x+3 = 25 + x - 1 - 10\sqrt{x-1}$$ $$2x+3 = x + 24 - 10\sqrt{x-1}$$ 6. **Step 4: Rearrange to isolate the radical:** $$2x + 3 - x - 24 = -10\sqrt{x-1}$$ $$x - 21 = -10\sqrt{x-1}$$ 7. **Step 5: Multiply both sides by -1:** $$-x + 21 = 10\sqrt{x-1}$$ 8. **Step 6: Square both sides again:** $$(-x + 21)^2 = (10\sqrt{x-1})^2$$ $$(-x + 21)^2 = 100(x-1)$$ 9. **Step 7: Expand and simplify:** $$x^2 - 42x + 441 = 100x - 100$$ 10. **Step 8: Bring all terms to one side:** $$x^2 - 42x + 441 - 100x + 100 = 0$$ $$x^2 - 142x + 541 = 0$$ 11. **Step 9: Solve quadratic using the quadratic formula:** $$x = \frac{142 \pm \sqrt{(-142)^2 - 4 \cdot 1 \cdot 541}}{2}$$ $$x = \frac{142 \pm \sqrt{20164 - 2164}}{2}$$ $$x = \frac{142 \pm \sqrt{18000}}{2}$$ 12. **Step 10: Simplify the square root:** $$\sqrt{18000} = \sqrt{100 \cdot 180} = 10\sqrt{180} = 10 \cdot 6\sqrt{5} = 60\sqrt{5}$$ 13. **Step 11: Final solutions:** $$x = \frac{142 \pm 60\sqrt{5}}{2} = 71 \pm 30\sqrt{5}$$ 14. **Step 12: Check for extraneous solutions by substituting back into the original equation. Only solutions that satisfy the original equation are valid. **Final answer:** $$x = 71 + 30\sqrt{5}$$ or $$x = 71 - 30\sqrt{5}$$ (verify which satisfy the original equation).