\begin{tabular}{l} Question 8 \\ Solve for \( x x^{2} \) \\ \( 7-\frac{18}{(6 x+4)^{-13}}+6=27 \) \\ Round to 3 decimal places \\ Rod your answer \\ \hline\end{tabular}

Solve the equation \( 7-\frac{18}{(6x+4)^{-13}}+6=27 \). Solve the equation by following steps: - step0: Solve for \(x\): \(7-\frac{18}{\left(6x+4\right)^{-13}}+6=27\) - step1: Find the domain: \(7-\frac{18}{\left(6x+4\right)^{-13}}+6=27,x\neq -\frac{2}{3}\) - step2: Simplify: \(13-18\left(6x+4\right)^{13}=27\) - step3: Move the expression to the left side: \(13-18\left(6x+4\right)^{13}-27=0\) - step4: Subtract the numbers: \(-14-18\left(6x+4\right)^{13}=0\) - step5: Rewrite the expression: \(-18\left(6x+4\right)^{13}=14\) - step6: Change the sign: \(18\left(6x+4\right)^{13}=-14\) - step7: Divide both sides: \(\frac{18\left(6x+4\right)^{13}}{18}=\frac{-14}{18}\) - step8: Divide the numbers: \(\left(6x+4\right)^{13}=-\frac{7}{9}\) - step9: Simplify the equation: \(\sqrt[13]{\left(6x+4\right)^{13}}=\sqrt[13]{-\frac{7}{9}}\) - step10: Calculate: \(6x+4=\sqrt[13]{-\frac{7}{9}}\) - step11: Simplify the root: \(6x+4=-\frac{\sqrt[13]{7\times 3^{11}}}{3}\) - step12: Move the constant to the right side: \(6x=-\frac{\sqrt[13]{7\times 3^{11}}}{3}-4\) - step13: Subtract the numbers: \(6x=-\frac{\sqrt[13]{7\times 3^{11}}+12}{3}\) - step14: Multiply by the reciprocal: \(6x\times \frac{1}{6}=-\frac{\sqrt[13]{7\times 3^{11}}+12}{3}\times \frac{1}{6}\) - step15: Multiply: \(x=-\frac{\sqrt[13]{7\times 3^{11}}+12}{18}\) - step16: Check if the solution is in the defined range: \(x=-\frac{\sqrt[13]{7\times 3^{11}}+12}{18},x\neq -\frac{2}{3}\) - step17: Find the intersection: \(x=-\frac{\sqrt[13]{7\times 3^{11}}+12}{18}\) The solution to the equation \(7-\frac{18}{(6x+4)^{-13}}+6=27\) is \(x \approx -0.830142\). Rounding to 3 decimal places, the answer is \(x \approx -0.830\).