Roots of Transformed Functions

Suppose an equation
$f(x)=0$
has solutions
$x=-3, \; 1, \; 2$
.
The function is then transformed to become
$f(3x-2)$
. What will be the solutions of the equation
$f(3x-2)=0$
?
The function actually tells us that
$f(-3)=f(1)=f(2)=0$
. We should be solving
$f(3x-2)=f(-3) \rightarrow 3x-2=-3 \rightarrow 3x=2-3=-1 \rightarrow x =- \frac{1}{3}$

$f(3x-2)=f(1) \rightarrow 3x-2=1 \rightarrow 3x=2+1=3 \rightarrow x =1$

$f(3x-2)=f(2) \rightarrow 3x-2=2 \rightarrow 3x=2+4=4 \rightarrow x = \frac{4}{3}$