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.

You're correct in dividing between $x\ge0$ and $x<0$.
Case $x\ge0$: the inequality is
$$
\frac{x^2-x-12}{x-3}\ge 2x
$$
that becomes
$$
\frac{x^2-x-12-2x^2+6x}{x-3}\ge0
$$
or
$$
\frac{x^2-5x+12}{x-3}\le0
$$
Since the numerator is positive (as the discriminant is negative), the solution set for this inequality is the interval $[0,3)$ (that is, $0\le x<3$), not the empty set. (This is where you got wrong.)
Case $x<0$: the inequality is
$$
\frac{x^2+x-12}{x-3}\ge 2x
$$
that becomes
$$
\frac{x^2+x-12-2x^2+6x}{x-3}\ge0
$$
or
$$
\frac{x^2-7x+12}{x-3}\le0
$$
or
$$
\frac{(x-3)(x-4)}{x-3}\le0
$$
which is satisfied for $x\le 4$ and $x\ne3$, but we have also the condition $x<0$, so the solution set is $(-\infty,0)$.
Putting together the two cases, we get
$$
(-\infty,3)
$$
