solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$. 
solve $\dfrac{x^2-|x|-12}{x-3}\geq 2x,\ \ x\in\mathbb{R}$.

options
$a.)\ -101<x<25\\
b.)\ [-\infty,3]\\
c.)\ x\leq 3\\
\color{green}{d.)\ x<3}\\
$
I tried ,
Case $1$ ,for $ \boxed{x\geq 0}\\
\dfrac{x^2-x-12}{x-3}\geq 2x\\
\implies \dfrac{x^2-5x+12}{x-3}\leq 0 \\
\implies x<3\\
x\in \emptyset $
Case $2$ ,for $\boxed{x< 0}\\
\dfrac{x^2+x-12}{x-3}\geq 2x\\
\implies \dfrac{(x-4)(x-3)}{x-3}\leq 0 \\
\implies x\leq 4\\
\implies x< 0\\
  $
But the answer given is option $d.)$
I look for a short and simple way.
I have studied maths up to $12$th grade.

Case $x>3$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \geq 2x(x-3) \Rightarrow x^2-x-12 \geq 2x^2-6x \\ \Rightarrow x^2-5x+12 \leq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So we reject this case. } $ 
Case $x<3$ :  


*

*Subcase $0<x<3$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2-x-12 \leq 2x(x-3) \Rightarrow x^2-x-12 \leq 2x^2-6x \\ \Rightarrow x^2-5x+12 \geq 0 $$ $\Delta=25-48<0 \text{ That means that } x^2-5x+12 \text{ has always the sign of } x^2 \\ \text{ so it is always positive. So  } \checkmark $ 

*Subcase $x<0$ : 
$$\frac{x^2-|x|-12}{x-3} \geq 2x \Rightarrow x^2+x-12 \leq 2x(x-3) \Rightarrow x^2+x-12 \leq 2x^2-6x \\ \Rightarrow x^2-7x+12 \geq 0 $$ $\Delta=49-48=1>0  \\ x_{1,2}=\frac{7 \pm 1}{2}= 3 \text{ or } 4 $ 
For $x \in (-\infty, 3] \cup [4, +\infty)$ we have that $x^2-7x+12 \geq 0$ but since we have that $x<0$ we conclude that in the case $x<0$ $x^2-7x+12 \geq 0$. 
In conclusion, it stands that $\frac{x^2-|x|-12}{x-3} \geq 2x$ for $x<3$. 
