But \(\displaystyle 3x+2\) COULD BE negative. You have to make sure that \(\displaystyle 3x+2 \) is NOT negative
Ok. So \(\displaystyle y = \sqrt{3x+2} \).
If, for example, \(\displaystyle x=4\) then \(\displaystyle y=\sqrt{3*4+2} = \sqrt {14}\) which is great because you can find the square root of 14, ie the square root of 14 exists (it's not rational but it exists). Therefore \(\displaystyle x=4\) is part of the domain.
If, for example,
\(\displaystyle x=-4\) then \(\displaystyle y=\sqrt{3*-4+2} = \sqrt {-10}\). This isn't good because you can't take the square root of -10, or any negative number for that matter (in the real number system anyway, but don't worry about that for now).
Therefore \(\displaystyle x=-4\) is NOT part of the domain. Because when you plug in -4 for x, the answer you get doesn't exist.
So,
the bit underneath the square root sign can NOT be negative, in other words it has to be positive or 0.
That is:
\(\displaystyle 3x+2 \geq0\). SO, solve this inequality for x and you'll have the domain.