If x  real show that value of the expression $\frac{x^2+34x-71}{x^2+2x-7}$ does not lie between 5 and 9

$$\begin{gathered} Let y=\frac{x^2+34x-71}{x^2+2x-7} \\ y\left(x^2+2x-7\right)=x^2+34x-71 \\ y{x}^2+2xy-7y=x^2+34x-71 \\ \Rightarrow y{x}^2+2xy-7y-x^2-34x+71=0 \\ \left(y-1\right)x^2+2\left(y-17\right)x+\left(-7y+71\right)=0 \\ Since x\in R, discriminent \ge 0 \\ \left\{ Compare with a{x}^2+bx+c\right\} \\ \Rightarrow b^2-4ac\ge 0. \\ \Rightarrow {\left[2\left[y-17\right]\right]}^2-4\left(y-1\right)\left(-7y+71\right)\ge 0 \\ {\left[2\left(y-17\right)\right]}^2-4\left(y-1\right)\left(-7y+71\right)\ge 0 \\ 4\left[y^2-34y+289\right]-4\left[-7y^2+78y-71\right]\ge 0 \\ 4\left[y^2-34y+289+7y^2-78y+71\right]\ge 0 \\ \Rightarrow 8y^2-112y+360\ge 0; 8\left[y^2-14y+45\right]\ge 0 \\ {y}^2-14y+45\ge 0; \left(y-5\right)\left(y-9\right)\ge 0 \\ y\le 5 or y\ge 9 \\ \left[If \alpha <\beta then \left(x-\alpha \right)\left(x-\beta \right)\ge 0\Rightarrow x\le \alpha or x\ge \beta \right] \\ y does not lie between 5 and 9 \end{gathered}$$