The range of the function $f\left(x\right)=2\sqrt{3x^2-4x+5}$ is

$f\left(x\right)$ is defined if $3x^2-4x+5\ge 0$
$$\begin{gathered} \Rightarrow 3\left[x^2-\frac{4}{3}x+\frac{5}{3}\right]\ge 0 \\ \Rightarrow \left[{\left(x-\frac{2}{3}\right)}^2+\frac{11}{9}\right]\ge 0 \end{gathered}$$
Which is true for all real $x$
$\therefore$ Domain $\left(f\right)=\left(-\infty ,\infty \right)$
Let, $y=2\sqrt{3x^2-4x+5}$
$$\begin{gathered} \Rightarrow \frac{y^2}{4}=3x^2-4x+5 \\ \mathrm{i}.\mathrm{e}. 3x^2-4x+\left(5-\frac{y^2}{4}\right)=0 \end{gathered}$$
For $x$ to be real, $16-12\left(5-\frac{y^2}{4}\right)\ge 0$

$\Rightarrow \frac{y}{2}\ge \sqrt{\frac{11}{3}}$
$\therefore$ Range of $y=\left[2\sqrt{\frac{11}{3}},\infty \right).$