6. Solve the equation $\sqrt{2 x^{2}+3 x+2}-\sqrt{2 x^{2}+3 x-5}=1$.
Solution: $\sqrt{2 x^{2}+3 x+2}-\sqrt{2 x^{2}+3 x-5}=1$, Domain $2 x^{2}+3 x-2 \geq 0$,
$2 x^{2}+3 x+2=t, \sqrt{t}-\sqrt{t-7}=1, \sqrt{t}=1+\sqrt{t-7}, t=1+t-7+2 \sqrt{t-7}$.

$6=2 \sqrt{t-7}, \sqrt{t-7}=3, t-7=9, t=16$.

$2 x^{2}+3 x+2=16,2 x^{2}+3 x-14=0, x_{1,2}=\frac{-3 \pm \sqrt{9+112}}{4}=\frac{-3 \pm 11}{4}=\left[\begin{array}{l}2 \\ -\frac{7}{2}\end{array}\right.$

Verification: $\sqrt{2 \cdot 4+6+2}-\sqrt{8+6-5}=1, \sqrt{2 \cdot \frac{49}{2^{2}}-\frac{21}{2}+2}-\sqrt{2 \cdot \frac{49}{2^{2}}-\frac{21}{2}-5}=1$,

Answer: $2,-\frac{7}{2}$.