Well, I'm not done yet. Am I on the right track?

$\displaystyle \sqrt{x^2-9x+20}\leq\sqrt{x-1}\leq\sqrt{x^2-13}$

$\displaystyle x^2-9x+20\leq{x-1}$

$\displaystyle x-1\leq{x^2-13}$

$\displaystyle x^2-10x+21\leq{0}$

$\displaystyle 0\leq{x^2-x-12}$

$\displaystyle (x-7)(x-3)\leq{0}$

$\displaystyle 0\leq{(x-4)(x+3)}$