jmc

Since $x^2-x-6=(x+2)(x-3),$ $x$ must satisfy $x\le -2$ or $x\ge 3.$ And since a square root is always nonnegative, $x$ must satisfy $2x-3\ge 0.$ Then $x\ge \frac{3}{2},$ so $x$ must satisfy $x\ge 3.$

Note that for $x\ge 3,$ both sides are nonnegative (and defined), so we can square both sides, to obtain the equivalent inequality $$x^2-x-6<4x^2-12x+9.$$This simplifies to $3x^2-11x+15>0.$ This inequality is satisfies for all real numbers, so the solution is $x\in \boxed{[3,\infty )}.$