Solve the equation with rational exponents. \[ \left(x^{2}-x-14\right)^{3 / 4}-5=3 \] Select the correct choice below and, il necessary, fill in the answer box to complete your choice. \( \begin{array}{l}\text { (i) A. The solution(s) is (are) } x=\square . \\ \text { (Simplity your answer. Use a comma to separate answers as needed.) } \\ \text { B. There is no solution. }\end{array} \)

Solve the equation \( (x^{2}-x-14)^{3 / 4}-5=3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left(x^{2}-x-14\right)^{\frac{3}{4}}-5=3\) - step1: Find the domain: \(\left(x^{2}-x-14\right)^{\frac{3}{4}}-5=3,x \in \left(-\infty,\frac{-\sqrt{57}+1}{2}\right]\cup \left[\frac{\sqrt{57}+1}{2},+\infty\right)\) - step2: Move the constant to the right side: \(\left(x^{2}-x-14\right)^{\frac{3}{4}}=3+5\) - step3: Add the numbers: \(\left(x^{2}-x-14\right)^{\frac{3}{4}}=8\) - step4: Raise both sides to the \(\frac{4}{3}\)-th power\(:\) \(\left(\left(x^{2}-x-14\right)^{\frac{3}{4}}\right)^{\frac{4}{3}}=8^{\frac{4}{3}}\) - step5: Evaluate the power: \(x^{2}-x-14=16\) - step6: Move the expression to the left side: \(x^{2}-x-30=0\) - step7: Factor the expression: \(\left(x-6\right)\left(x+5\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-6=0\\&x+5=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=6\\&x=-5\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=6\\&x=-5\end{align},x \in \left(-\infty,\frac{-\sqrt{57}+1}{2}\right]\cup \left[\frac{\sqrt{57}+1}{2},+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=6\\&x=-5\end{align}\) - step12: Check the solution: \(\begin{align}&x=6\\&x=-5\end{align}\) - step13: Rewrite: \(x_{1}=-5,x_{2}=6\) The solutions to the equation \((x^{2}-x-14)^{3 / 4}-5=3\) are \(x=-5\) and \(x=6\). Therefore, the correct choice is: (i) A. The solution(s) is (are) \(x=-5, 6\).