√(6x2+27x+4)=x+2, squaring:
6x2+27x+4=x2+4x+4,
5x2+23x=0=x(5x+23), so x=0 or -23/5, but we need to check for extraneous solutions.
When x=0: √4-2=0 OK.
When x=-23/5 we have √(169/25)=13/5.
This is the positive square root and 13/5-2=⅗≠-23/5 so this is an extraneous solution.
(The negative square root does satisfy the original equation: -13/5-2=-23/5 but we cannot use this because the original equation explicitly specifies the positive square root.)