How to solve $\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$ where $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$? 
Let $t=\sqrt{2}-\sqrt{3}-\sqrt{5}+\sqrt{6}$
then fine the $x$:
$$\frac{2}{3}tx+\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x}=t+\sqrt{2}+\sqrt{6}$$


since : $$\sqrt{2x-3} \to 2x-3 \geq 0 \to x\geq\dfrac{3}{2}$$
$$\sqrt{2x-1} \to 2x-1 \geq 0 \to x\geq\dfrac{1}{2}$$
$$\sqrt{3-x} \to 3-x \geq 0 \to x\leq 3$$
So :
$$ \dfrac{3}{2} \leq x \leq 3$$
Now what ?

if you see $d(d(\sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x})/dx)/dx$, you will see that is less than zero from $\frac{3}{2}$ to $3$. So $f = \frac{2}{3}tx + \sqrt{2x-3}+\sqrt{2x-1}+2\sqrt{3-x} - t - \sqrt{2} - \sqrt{6}$ is convex upward. At $x = \frac{3}{2}$ is zero, at $x = 3$ is zero too. So we have no other roots.
