Theory of Equations

$\begin{aligned} & x^2 +4\left(\frac{x}{x-2}\right)^2 -45 = 0 \\ & x^2+\left(\frac{2x}{x-2}\right)^2 -45 = 0 \\ & \left(x+\frac{2x}{x-2}\right)^2 - 2.(2x)\left(\frac{x}{x-2}\right) -45 = 0 \\ & \left(\frac{x^2}{x-2}\right)^2 -4\left(\frac{x^2}{x-2}\right) -45 = 0 \\& \left(\frac{x^2}{x-2}\right)^2 - 9\left(\frac{x^2}{x-2}\right) +5\left(\frac{x^2}{x-2}\right) -45 = 0 \\ & \frac{x^2}{x-2}\left(\frac{x^2}{x-2} -9\right) + 5\left(\frac{x^2}{x-2} -9\right) = 0 \\ & \left(\frac{x^2}{x-2}-9\right)\left(\frac{x^2}{x-2}+5\right) = 0 \end{aligned}$

Either $\begin{aligned} & \left(\frac{x^2}{x-2}-9\right) = 0 \\ & \Rightarrow x^2-9x +18 = 0 \implies (x-3)(x-6)=0 \\ & x = 6\space \text{or} x = 3 \space \text{(restricted)} \end{aligned}$

0r $\begin{aligned} & \left(\frac{x^2}{x-2}+5\right) = 0 \\ & \Rightarrow x^2 +5x-10= 0 \\ & \implies x = \frac{-5±\sqrt {45}}{2} \\ & \implies x = \frac{\sqrt{45}-5}{2} \approx 0.85410..\\ & \implies x= \frac{-\sqrt{45} -5}{2} \approx - 5.85410..\end{aligned}$ .

Thus we have $x = 6, 0.58410... ,-5.85410$ Taking $x=6$ and substituting in $\frac{(x-2)^2(x+3)}{2x-3} = \frac{16\times9}{9} \Rightarrow \boxed{16}$

@Naren Bhandari , @Beatriz Sampaio how do u solve this question? Could u pls post the solution