Factorise 8+2x-3x²=(2-x)(4+3x). This expression must be positive to take its square root, so x≤2 and x≥-4/3.
Therefore the domain is -4/3≤x≤2 or [-4/3,2].
The range starts at 0 because when x≤2 or x≥-4/3, the expression is 0.
We need to find the maximum of the expression:
8+2x-3x²=8-(-2x+3x²)=
8-3(-⅔x+x²)=8-3(1/9-⅔x+x²-1/9)=
8-3((⅓-x)²-1/9)=8-3(⅓-x)²+⅓=8⅓-3(⅓-x)².
This has a maximum value of 8⅓ (25/3), so the range is [0,√(25/3)]=[0,5√3/3].