6. Solve the equation $\sqrt{2 x^{2}+3 x-5}-\sqrt{2 x^{2}+3 x-8}=1$.
Solution: $\sqrt{2 x^{2}+3 x-5}-\sqrt{2 x^{2}+3 x-8}=1,\left\{\begin{array}{l}2 x^{2}+3 x-5 \geq 0, \\ 2 x^{2}+3 x-8 \geq 0,\end{array}\left\{\begin{array}{l}x \leq-\frac{3-\sqrt{73}}{4}, \\ x \geq-\frac{3+\sqrt{73}}{4}\end{array}\right.\right.$,
$\sqrt{t}=1+\sqrt{t-3} ; t=1+2 \sqrt{t-3}+t-3 ; \sqrt{t-3}=1 ; t-3=1, t=4$.

$2 x^{2}+3 x-5=4,2 x^{2}+3 x-9=0, x_{1,2}=\frac{-3 \pm \sqrt{9+72}}{4}=\frac{-3 \pm 9}{4}=\left[\begin{array}{l}\frac{3}{2} \\ -3 .\end{array}\right.$

Verification: $\sqrt{18-9-5}-\sqrt{18-9-8}=1, \sqrt{2 \cdot \frac{9}{4}+\frac{9}{2}-5}-\sqrt{2 \cdot \frac{9}{4}+\frac{9}{2}-8}=1,2-1=1$