### Instruction: Find all solutions to the equation $\sqrt{3x+6}=x+2$.  If there are multiple solutions, order them from least to greatest, separated by comma(s).### Response: To solve this equation, I need to isolate x on one side and then check if the solutions are valid or extraneous.
I can start by squaring both sides to get rid of the square root on the left.
This gives me $3x+6=(x+2)^2$.
Expanding the right side, I get $3x+6=x^2+4x+4$.
Subtracting $3x+6$ from both sides, I get $0=x^2+x-2$.
This is a quadratic equation that I can factor as $0=(x+2)(x-1)$.
Using the zero product property, I get $x+2=0$ or $x-1=0$.
Solving for x, I get $x=-2$ or $x=1$.
However, I need to check if these solutions are valid by plugging them back into the original equation.
If I plug in $x=-2$, I get $\sqrt{3(-2)+6}=-2+2$, which simplifies to $\sqrt{0}=0$, which is true.
If I plug in $x=1$, I get $\sqrt{3(1)+6}=1+2$, which simplifies to $\sqrt{9}=3$, which is also true.
Therefore, both $x=-2$ and $x=1$ are valid solutions to the equation, and they are already ordered from least to greatest.