jmc

Because $(x-4{)}^2$ is always nonnegative, we can safety multiply both sides of the inequality by $(x-4{)}^2$ without changing the direction of the inequality, with the caveat that we cannot have $x=4$: $$\begin{array}{rl}x(x+1)& \ge 12(x-4{)}^2\\ 0& \ge 11x^2-97x+192.\end{array}$$This quadratic factors as $$0\ge (x-3)(11x-64),$$which holds if and only if $3\le x\le \frac{64}{11}.$ However, since $x\ne 4,$ the solutions to the original inequality are given by $$x\in \boxed{[3,4)\cup \left(4,\frac{64}{11}\right]}.$$