Which equation has the solutions x=-3+or- square root 3i over 2?

If you set the whole thing equal to x as one solution using both the positive and negative of the square root you get x = [-3 +/- (sqrt3)*i]/2.  Multiply both sides by 2 to get 2x = -3+/- (sqrt3)*i.  Next add 3 to both sides to get (2x+3) = +/-(sqrt3)*i.  Squaring both sides not only takes care of the square root sign, it also eliminates the need for the +/- since the +/- came in by taking the square root in the first place.  Now you have this as your equation:
(2x+3)^2 = ((sqrt3)*i)^2.  Squaring both sides gives you 4x^2 + 12x + 9 =3i^2.
i^2 is = to -1, so NOW we have 4x^2 + 12x + 9 = 3(-1) or 4x^2+12x+9=-3.  Add 3 to both sides to get 4x^2 + 12x + 12 = 0 as your equation.  You didn't list the choices, but I'm guessing this is one of them!