Non-negative root of ${\left(\frac{3x-1}{2x+3}\right)}^2-5\left(\frac{3x-1}{2x+3}\right)+4=0$ is _____.

The correct option is C.$4$

Let $\frac{3x-1}{2x+3}=y$

The equation ${\left(\frac{3x-1}{2x+3}\right)}^2-5\left(\frac{3x-1}{2x+3}\right)+4=0$ reduces to:
$y^2-5y+4=0$

$\Rightarrow y^2-4y-y+4=0$

$\Rightarrow y(y-4)-1(y-4)=0$
$\Rightarrow (y-1)(y-4)=0$
$\Rightarrow y=1,4$
Now, $\frac{3x-1}{2x+3}=1$
$\Rightarrow 3x-1=2x+3$

$\Rightarrow 3x-2x=1+3$
$\Rightarrow x=4$
and

$\frac{3x-1}{2x+3}=4$
$\Rightarrow 3x-1=8x+12$

$\Rightarrow 3x-8x=1+12$

$\Rightarrow -5x=13$

$\Rightarrow x=\frac{-13}{5}$ is negative.

Thus, only $x=4$ is the required root.