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!