Complete the square and write in standard form for $3x^2+3x+2y=0$ 
Standard forms: $y-b=A(x-a)^2$ or $x-a=A(y-b)^2$
   $3x^2+3x+2y=0$  

I honestly do not know how to start this problem. I have tried a lot of things and obviously not the right one. Can someone explain to me the first step and nothing more and I will edit with my new discoveries. Thanks!

Since you have a factor with $x^2$, this anticipates the form you will be aiming for is
$$y-b=A(x-a)^2$$
So let's look at your eqn.:
$$3x^2+3x+2y=0$$
We need to produce a perfect square with $3x^2+3x$. So, we can do the old completing the square trick:
$$3x^2+3x=3(x^2+x)$$
$$3x^2+3x=3(x^2+2\frac 1 2 x)$$
$$3x^2+3x=3\left[x^2+2\frac 1 2 x+\left( \frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left[\left( x+\frac1 2 \right)^2 -\left( \frac1 2 \right) ^2\right]$$
$$3x^2+3x=3\left( x+\frac1 2 \right)^2- \frac3 4$$
Can you move on?
