"Charges of +3.0mC and -2.0mC are placed 20.0 cm apart. Where is the net electric field equal to 0?"

The best I can come up with is (.000003 C)/d1^2 = -(.000002 C)/d2^2 but that still leaves me with two variables...

When the charges are in the opposite in nature, neutral point will be out side the two charges. If d is the distance between the charges, and x is the distance of the neutral point from the smaller charge, then
k*q1/d^2 = k*q2/(d+x)^2.
Now solve for x.

This is actually not such a bad problem to solve. We know the distance between the two unlike charges is 20 cm, so let's make it easy on ourselves by assuming that the positive 3 mC of charge is at the origin of an x-y coordinate system and that the negative charge of -2 mC is 0.2 m down the x axis. Now let's place a "test" charge at some distance down the x axis such that the sum of the E fields from both differing charges is zero at some distance. This should make sense as the positive charge is greater in quantity so its E field is stronger, but further away from the point where the negative charge is located, which is lesser in quantity (so its E field is not as strong), but is closer to the point where we are finding the two E fields to be equal and opposite in strength. So there must be a distance down the x axis that has an E field from the positive charge that is equal and opposite to that from the negative charge. So one is stronger, but further away, so its influence at some distance will be the same (but opposite) to that of the weaker E field that is closer to the point where the two are equal and opposite in strength.

So, since we are going to set the sum of each E field function for each charge equal to zero, we can divide out the constant k that is usually in front of the function, leaving something it in the form of q/d^2 as the general form of the function we will use for each charge emitting or receiving an E field. OK, let's start:

q1/(d+0.2)^2 + q2/d^2 = 0

where q1 is the positive charge at the origin and q2 is the negative charge 0.2 m down the x axis, and d is the distance to the place where the net E field is zero with respect to the negative charge, and d+0.2m away from the positive charge, also down the x axis for convenience.

Now let's put in the values for the charge, you have them written as 3 mC and -2 mC, but you wrote them in your equation as 3 micro Coulombs and -2 micro Coulombs, so I don't know if they are really milli or micro Coulombs of charge you are talking about, so I will go with what you have written, the milli-Coulombs, and if instead you need the distance using micro Coulombs of charge, just do the same thing I'm about to do, except replace the milli-Coulombs with the micro Coulombs and you'll get the answer you want. In fact I'll do both, and then you'll have the answer for both micro and milli-Coulombs of charge. Anyway, let's plug in the charge values and sign.

0.003/(d+0.2)^2 - 0.002/d^2 = 0

The reason why the positive charge is 0.2m further away from the place where the two E fields should sum to zero is because it is at the origin and the negative charge is 20 cm down the x axis, so the negative charge is a distance d away from this point while the positive charge is a distance of d + 20cm away from this point where the two E fields sum to zero. So the place where the net E field is going to be zero is closer to the negative charge and further from the positive charge, so I'll give you the distances from each charge down the x axis where the net E field is zero.

0.003/(d+0.2)^2 = 0.002/d^2
0.003/0.002 = [(d+0.2)/d]^2
1.5 = [1+0.2/d]^2
1.224745 = 1+0.2/d
0.224745 = 0.2/d
0.224745/0.2 = 1/d

d = 0.2/0.224745 = 0.889898 meters away from the negative charge which is 20 cm away from the origin and the positive charge, so the location where the net E field is zero when the positive charge is placed at the origin and the negative charge is placed 20cm down the x axis is approximately 1.09 meters away from the origin down the x axis where the positive charge is located at the origin, and is about 89 cm away from the negative charge that is 20 cm down the x axis away from the origin.

Now if we want to know the distance using 3 micro Coulombs and -2 micro Coulombs it will be:

q1/(d+0.2)^2 + q2/d^2 = 0
0.000003/(d+0.2)^2 = 0.000002/d^2
1.5 = [(d+0.2)/d]^2

This is going to give us the same answer as using the milli-Coulombs because the ratio of the charge is the same, so as long as the charge ratio is always the same then the distance where the net E field is zero will always be 1.09 meters from the origin (or from the positive charge) and is about 89 cm away from where the negative charge is located.

This has to be such that the distance away from the positive charge is 1.09 meters and 89 cm away from the negative charge. If those two distances are not met, the net E field will not be zero, so be sure to carefully word your answer such that it is clear that the distance where the net E field is zero is 1.09 meters away from the positive charge and is 89 cm away from the negative charge; as long as that is true for both charges then the net E field will be zero.

Hope that helps some,
Many Smiles,
Craig