√(3x+3)+√(x+2)=5, square both sides:
3x+3+2√[(3x+3)(x+2)]+x+2=25,
4x+5+2√[(3x+3)(x+2)]=25,
2√[(3x+3)(x+2)]=20-4x, squaring both sides:
4(3x2+9x+6)=400-160x+16x2,
12x2+36x+24=400-160x+16x2,
4x2-196x+376=0, divide through by 4:
x2-49x+94=0=(x-2)(x-47).
There appear to be two solutions for x: 2 and 47, but we need to test them using the original equations.
x=2: √9+√4=3+2=5, so x=2 is a solution.
x=47: √144+√49=12+7≠5, so x=47 is not a solution.