The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\  2x-3   = -(2x-3)$ when $2x-3$ $\geqslant$  $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$  $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is  presented in interval notation (-$\infty$ , $\frac{3}{2}$].
Any help is appreciated.

$$\left | 2x-3 \right | = -(2x-3)$$
$let$,   $t= 2x-3$
$$\left | t \right | = -t$$ 
$$t=<0$$
$$2x-3=<0$$
$$x \in \left(-\infty, \frac{3}{2}\right]$$
