6. Solve the equation $\sqrt{3 x^{2}+5 x+7}-\sqrt{3 x^{2}+5 x+2}=1$.
Solution: Domain of definition $3 x^{2}+5 x+2 \geq 0 . x_{1,2}=\frac{-5 \pm \sqrt{25-24}}{6}=\frac{-5 \pm 1}{6}=\left[\begin{array}{l}\frac{2}{3}, \\ -1\end{array}\right.$

$t=3 x^{2}+5 x+7, \quad \sqrt{t}=1+\sqrt{t-5} ; \quad t=1+2 \sqrt{t-5}+t-5$;

$4=2 \sqrt{t-5} ; t-5=4 ; t=9 ; 3 x^{2}+5 x+7=9 ; 3 x^{2}+5 x-2=0$;

![](https://cdn.mathpix.com/cropped/2024_05_06_8eeeb8d6ace591217257g-04.jpg?height=193&width=547&top_left_y=2639&top_left_x=1414)
$x=\frac{-5 \pm \sqrt{25-24}}{6}=\frac{-6 \pm \sqrt{49}}{6}=\frac{-5 \pm 7}{6}=\left[\begin{array}{ll}\frac{1}{3}, & \sqrt{9}-\sqrt{4}=1-\text { correct, } \\ -2, & \sqrt{12-10+7}-\sqrt{12-10+2}=1,3-2=1,\end{array}\right.$

Verification: $\sqrt{12-10+7}-\sqrt{12-10+2}=1, \sqrt{\frac{1}{3}+\frac{5}{3}+7}-\sqrt{\frac{1}{3}+\frac{5}{3}+2}=1$. Answer: $-2 ; \frac{1}{3}$.