\sqrt{3x+1} - 4 + 3x^{2} -14x -5 = \sqrt{6-x} -1\rightarrow (x- 5)(\frac{1}{\sqrt{3x+1}+ 4}+ (3x+1) + \frac{1}{\sqrt{6 - x}+ 1}=0\rightarrow x=5
$\sqrt{3x+1}$ +3$x^{2}$ -14x -8= $\sqrt{6-x}$$\rightarrow$$\sqrt{3x+1}$ - 4 + 3$x^{2}$ -14x -5 = $\sqrt{6-x}$ -1$\rightarrow (x- 5)$$(\frac{1}{\sqrt{3x+1}+ 4}$+ 3x+1 + $\frac{1}{\sqrt{6 - x}+ 1}$=0 ( nhân liên hợp)$\rightarrow $x=5