1. Stated problem: Solve the equation $$\sqrt[4]{x-6} + \sqrt{x^2+36} = 0$$. 2. Analyze the terms: The fourth root $$\sqrt[4]{x-6}$$ is defined only if $$x-6 \geq 0$$, so $$x \geq 6$$. 3. Note that both $$\sqrt[4]{x-6}$$ and $$\sqrt{x^2+36}$$ are non-negative for all valid $$x$$, because - $$\sqrt[4]{x-6} \geq 0$$ for $$x \geq 6$$, - $$\sqrt{x^2+36} \geq 0$$ for all real $$x$$. 4. Since both expressions are non-negative, their sum can be zero only if each is individually zero: $$\sqrt[4]{x-6} = 0$$ and $$\sqrt{x^2+36} = 0$$. 5. Solve $$\sqrt[4]{x-6} = 0 \Rightarrow x-6=0 \Rightarrow x=6$$. 6. Solve $$\sqrt{x^2+36} = 0 \Rightarrow x^2 + 36 = 0 \Rightarrow x^2 = -36$$ which has no real solutions. 7. Because $$\sqrt{x^2+36}$$ is never zero for real $$x$$, there are no real solutions to the equation. Final answer: No real solutions exist.