6.167. $\sqrt{x^{2}+x+4}+\sqrt{x^{2}+x+1}=\sqrt{2 x^{2}+2 x+9}$.
Solution.

From the condition $\sqrt{x^{2}+x+4}+\sqrt{x^{2}+x+1}=\sqrt{2\left(x^{2}+x\right)+9}$.

Let $x^{2}+x=y$. The equation in terms of $y$ becomes

$$
\sqrt{y+4}+\sqrt{y+1}=\sqrt{2 y+9}
$$

Squaring both sides of the equation, we get

$y+4+2 \sqrt{(y+4)(y+1)}+y+1=2 y+9 \Leftrightarrow \sqrt{(y+4)(y+1)}=2 \Leftrightarrow$ $\Leftrightarrow y^{2}+5 y+4=4, y^{2}+5 y=0$, from which $y_{1}=0, y_{2}=-5$.

Then $x^{2}+x=0$ or $x^{2}+x=-5$. Solving each of the obtained equations, we find $x_{1}=0, x_{2}=-1$; the second equation has no solutions, since $D<0$.

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