Solution set of the inequations $x^2-4x\le 12$ and $x^2-2x\ge 15$  taken together is

The given inequations are $x^2-4x\le 12\Rightarrow x^2-4x-12\le 0$

$\Rightarrow (x-6)(x+2)\le 0\Rightarrow x\in \left[-2,6\right]$    ….. ( 1 )

And  $x^2-2x\ge 15\Rightarrow x^2-2x-15\ge 0$

$\Rightarrow (x-5)(x+3)\ge 0$

$\Rightarrow x\le -3orx\ge 5$

$\Rightarrow x\in \left(-\infty ,-3\right]\cup \left[5,\infty \right)$   …… ( 2 )

$\therefore$   Solution set of the inequations is $\left[-2,6\right]\cap \left(-\infty ,-3\right]\cup \left[5,\infty \right)=\left[5,6\right]$