We have the expression:
[tex]\sqrt[]{-3x-2}=x+2[/tex]
The square root, to be defined in the domain of real numbers, has to have a 0 or positive argument. Then, -3x-2 has to be greater or equal than 0.
We can write:
[tex]\begin{gathered} -3x-2\ge0 \\ -3x\ge2 \\ x\le-\frac{2}{3} \end{gathered}[/tex]
Then, solutions for x that are greater than -2/3 are not valid.
We are left with x=-6 and x=-1.
We can test both:
[tex]\begin{gathered} x=-6\Rightarrow\sqrt[]{-3(-6)-2}=-6+2 \\ \sqrt[]{18-2}=-4 \\ \sqrt[]{16}=-4 \\ 16=(-4)^2 \\ 16=16 \end{gathered}[/tex]
x=-6 is a valid solution.
We now test the other solution:
[tex]\begin{gathered} x=-1\longrightarrow\sqrt[]{-3(-1)-2}=-1+2 \\ \sqrt[]{3-2}=1 \\ \sqrt[]{1}=1 \\ 1=1 \end{gathered}[/tex]
x=-1 is also a valid solution.
