6.168. $\sqrt{3 x^{2}-2 x+15}+\sqrt{3 x^{2}-2 x+8}=7$.
## Solution.

Let $3 x^{2}-2 x=y$. With respect to $y$, the equation takes the form $\sqrt{y+15}+\sqrt{y+8}=7$. Squaring both sides of the equation, we get $y+15+2 \sqrt{(y+15)(y+8)}+y+8=49 \Leftrightarrow \sqrt{(y+15)(y+8)}=13-y$, where $13-y \geq 0, y \leq 13$. From this, $(y+15)(y+8)=(13-y)^{2}, y=1$.

Then $3 x^{2}-2 x=1,3 x^{2}-2 x-1=0$, from which $x_{1}=-\frac{1}{3}, x_{2}=1$.

Answer: $x_{1}=-\frac{1}{3}, x_{2}=1$.