Logging inequalities 1

Nice question ... My solution is a bit lengthy but I tried not to skip any step!! ;-)
First we have to make a set in which we have to solve for x:
$(1.)$ Since square root is defined only for non negative quantities, we get $x^2-x-6\geq 0$ or $\color{#D61F06}{x\in (-\infty,-2] \cup [3,\infty)}$ .
$(2.)$ RHS of the inequality must be a non negative since square root of any quantity cannot be less than a negative, so we get $2x-3\geq0$ or $\color{#D61F06}{x\in[\dfrac{3}{2},\infty)}$ .
From $(1.)$ and $(2.)$ we have to find solutions for our inequality in $\color{#0C6AC7}{[3,\infty)}$ .

Squaring both sides of the inequality(since both sides are non negative) to get: $x^2-x-6<4x^2-12x+9$ $\Rightarrow 3x^2-11x+15>0$ We note that discriminate of $3x^2-11x+15$ is negative and leading coefficient is positive and hence inequality is satisfied $\forall x\in\mathbb{R}$ . But we have to be cautious here before claiming the answer since we have to look only in the set we have previously found. $\huge\boxed{\color{#007fff}{\therefore x \in [3,\infty)}}$