Solve the inequation: $\;$ $C^{x+1}_{x-1} > \dfrac{3}{2}$, $\;$ $x \in N$
$C^{x+1}_{x-1} > \dfrac{3}{2}$
i.e. $\;$ $\dfrac{\left(x+1\right)!}{\left(x-1\right)! \left(x+1-x+1\right)!} > \dfrac{3}{2}$
i.e. $\;$ $\dfrac{\left(x+1\right) x \left(x-1\right)!}{\left(x-1\right)! 2!} > \dfrac{3}{2}$
i.e. $\;$ $\dfrac{\left(x+1\right) x}{2} > \dfrac{3}{2}$
i.e. $\;$ $x^2 + x > 3$
i.e. $\;$ $x^2 + x - 3 > 0$ $\;\;\; \cdots \; (1)$
Solving the equation $\;$ $x^2 + x - 3 = 0$ $\;\;\; \cdots (2)$ gives
$x = \dfrac{-1 \pm \sqrt{1 - 4 \times 1 \times \left(-3\right)}}{2 \times 1} = \dfrac{-1 \pm \sqrt{13}}{2}$
i.e. $\;$ $x = \dfrac{-1 + \sqrt{13}}{2} = 1.302$ $\;$ or $\;$ $x = \dfrac{-1 - \sqrt{13}}{2}$
$\because \;$ $x \in N$, $\;$ the possible solution to $(1)$ is $\;$ $x \geq 2$