6.036. $\sqrt{3 x+7}-\sqrt{x+1}=2$.

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6.036. $\sqrt{3 x+7}-\sqrt{x+1}=2$.
Solution.

Domain of definition: $\left\{\begin{array}{l}3 x+7 \geq 0, \\ x+1 \geq 0\end{array} \Leftrightarrow x \geq-1\right.$.

By squaring both sides of the equation, we get

$$
\begin{aligned}
& 3 x+7-2 \sqrt{(3 x+7)(x+1)}+x+1=4 \Leftrightarrow 4 x+4=2 \sqrt{(3 x+7)(x+1)} \Leftrightarrow \\
& \Leftrightarrow 2(x+1)=\sqrt{(3 x+7)(x+1)} \Rightarrow 4(x+1)^{2}=(3 x+7)(x+1) \Leftrightarrow \\
& \Leftrightarrow 4(x+1)^{2}-(3 x+7)(x+1)=0 \Leftrightarrow(x+1)(4 x+4-3 x-7)=0 \Leftrightarrow \\
& \Leftrightarrow(x+1)(x-3)=0 \text { or } x+1=0, x_{1}=-1 \text { or } x-3=0, x_{2}=3 .
\end{aligned}
$$

By direct verification, we confirm that these are the roots of the original equation.

Answer: $x_{1}=-1, x_{2}=3$.