1. **State the problem:** Solve the equation $$x = \sqrt{x-1} - 2$$ for $$x > 2$$. 2. **Rewrite the equation:** Add 2 to both sides to isolate the square root: $$x + 2 = \sqrt{x-1}$$ 3. **Square both sides** to eliminate the square root: $$ (x + 2)^2 = (\sqrt{x-1})^2 $$ $$ (x + 2)^2 = x - 1 $$ 4. **Expand the left side:** $$ (x + 2)^2 = x^2 + 4x + 4 $$ 5. **Set up the quadratic equation:** $$ x^2 + 4x + 4 = x - 1 $$ 6. **Bring all terms to one side:** $$ x^2 + 4x + 4 - x + 1 = 0 $$ $$ x^2 + 3x + 5 = 0 $$ 7. **Solve the quadratic equation:** Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$a=1$$, $$b=3$$, $$c=5$$. Calculate the discriminant: $$ \Delta = 3^2 - 4 \times 1 \times 5 = 9 - 20 = -11 $$ Since $$\Delta < 0$$, there are no real solutions. 8. **Check the domain:** The original equation requires $$x > 2$$ and the square root is defined for $$x-1 \geq 0$$, so $$x \geq 1$$. 9. **Conclusion:** No real solutions satisfy the equation for $$x > 2$$. **Final answer:** No real solutions for $$x > 2$$.