$6.285 \frac{x^{2}}{\sqrt{5 x+4}}+\sqrt{5 x+4}=\frac{4}{3} x+2$.
Solution. Let's square both sides:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 5 x + 4 > 0 , } \\
{ 2 x + 3 > 0 , } \\
{ \frac { x ^ { 4 } } { 5 x + 4 } + 5 x + 4 + 2 x ^ { 2 } = \frac { 4 } { 9 } ( 2 x + 3 ) ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
5 x+4>0, \\
\frac{9 x^{4}}{5 x+4}+2 x^{2}-3 x=0
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ 5 x + 4 > 0 , } \\
{ [ \begin{array} { l } 
{ x = 0 , } \\
{ \frac { 9 x ^ { 3 } } { 5 x + 4 } }
\end{array} + ( 2 x - 3 ) = 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0, \\
\left\{\begin{array}{l}
5 x+4>0, \\
9 x^{3}+10 x^{2}-7 x-12=0
\end{array}\right.
\end{array} \Leftrightarrow\right.\right.
\end{aligned}
$$

$\Leftrightarrow\left[\begin{array}{l}x=0, \\ \left\{\begin{array}{l}5 x+4>0, \\ (x-1)\left(9 x^{2}+19 x+12\right)=0\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}x=0, \\ x=1 .\end{array}\right.\right.$

Answer: $\quad x_{1}=0, x_{2}=1$.