1. **State the problem:** Solve the equation $$\sqrt{2x+2} = -2 + \sqrt{6x-6}$$. 2. **Important note:** The square root function $$\sqrt{y}$$ is defined only for $$y \geq 0$$ and always returns a non-negative value. Therefore, the left side $$\sqrt{2x+2} \geq 0$$. The right side is $$-2 + \sqrt{6x-6}$$, which must also be non-negative for equality to hold. 3. **Check domain restrictions:** - For $$\sqrt{2x+2}$$, require $$2x+2 \geq 0 \Rightarrow x \geq -1$$. - For $$\sqrt{6x-6}$$, require $$6x-6 \geq 0 \Rightarrow x \geq 1$$. So the domain is $$x \geq 1$$. 4. **Analyze the right side:** Since $$\sqrt{6x-6} \geq 0$$, then $$-2 + \sqrt{6x-6} \geq -2$$. For the right side to be non-negative, we need: $$-2 + \sqrt{6x-6} \geq 0 \Rightarrow \sqrt{6x-6} \geq 2$$ Square both sides: $$6x - 6 \geq 4 \Rightarrow 6x \geq 10 \Rightarrow x \geq \frac{10}{6} = \frac{5}{3} \approx 1.666...$$ 5. **Update domain:** $$x \geq \frac{5}{3}$$. 6. **Isolate the square roots:** $$\sqrt{2x+2} + 2 = \sqrt{6x-6}$$ 7. **Square both sides:** $$\left(\sqrt{2x+2} + 2\right)^2 = (\sqrt{6x-6})^2$$ $$ (\sqrt{2x+2})^2 + 2 \times 2 \times \sqrt{2x+2} + 2^2 = 6x - 6$$ $$ 2x + 2 + 4\sqrt{2x+2} + 4 = 6x - 6$$ 8. **Simplify:** $$ 2x + 6 + 4\sqrt{2x+2} = 6x - 6$$ 9. **Isolate the square root term:** $$4\sqrt{2x+2} = 6x - 6 - 2x - 6 = 4x - 12$$ 10. **Divide both sides by 4:** $$\cancel{4}\sqrt{2x+2} = \cancel{4}(x - 3) \Rightarrow \sqrt{2x+2} = x - 3$$ 11. **Since $$\sqrt{2x+2} \geq 0$$, we require $$x - 3 \geq 0 \Rightarrow x \geq 3$$**. 12. **Square both sides again:** $$ (\sqrt{2x+2})^2 = (x - 3)^2$$ $$ 2x + 2 = (x - 3)^2 = x^2 - 6x + 9$$ 13. **Bring all terms to one side:** $$ 0 = x^2 - 6x + 9 - 2x - 2 = x^2 - 8x + 7$$ 14. **Solve quadratic equation:** $$x^2 - 8x + 7 = 0$$ Use quadratic formula: $$x = \frac{8 \pm \sqrt{64 - 28}}{2} = \frac{8 \pm \sqrt{36}}{2} = \frac{8 \pm 6}{2}$$ 15. **Calculate roots:** - $$x = \frac{8 + 6}{2} = 7$$ - $$x = \frac{8 - 6}{2} = 1$$ 16. **Check domain and original equation:** - $$x=1$$ is less than 3, so discard. - $$x=7$$ satisfies $$x \geq 3$$. 17. **Verify $$x=7$$ in original equation:** Left side: $$\sqrt{2(7)+2} = \sqrt{14 + 2} = \sqrt{16} = 4$$ Right side: $$-2 + \sqrt{6(7) - 6} = -2 + \sqrt{42 - 6} = -2 + \sqrt{36} = -2 + 6 = 4$$ Both sides equal 4, so $$x=7$$ is a valid solution. **Final answer:** $$\boxed{7}$$