\(-\dfrac{1}{3}\le x\le6\)
\(\sqrt{3x+1}-4-\left(\sqrt{6-x}-1\right)+3x^2-14x-5=0\)
\(\Leftrightarrow\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1\right)=0\)
do \(x\ge\dfrac{-1}{3}\Rightarrow3x+1\ge0\)
\(\dfrac{3}{\sqrt{3x+1}}+\dfrac{1}{\sqrt{6-x}+1}+3x-1>0\)
\(\Rightarrow x=5\)
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