Three vertices of a parallelogram have coordinates (-2,2),(1,6) and (4,3). Find all possible coordinates of the fourth vertex. Do I use the distance formula for the two points and use them to add to each other to get two parallel sides?
If three points are $P,Q,R$ then $R+(P-Q)$ gives a fourth vertex for a parallelogram. So pick the ordered pair $(P,Q)$ in all six ways, and that gives three parallelograms.

There are actually only three such paralellograms, some using my description being repeats (same vertices). So if $P,Q,R$ are the three given points (which are not collinear) the three parallelograms are those formed by using the given three vertices along with any one of the three choices $$P+Q-R,\ P+R-Q,\ Q+R-P$$ as the fourth vertex of the parallelogram.

Added note: In each case the subtracted point winds up being diagonally opposite the constructed point in that parallelogram. For example, if $X=P+Q-R,$ then also $X-P=Q-R$ as expected in a parallelogram labeled going around say counterclockwise in the order $X,P,R,Q.$ The equality of the vectors $X-P$ and $Q-R$ means they are parallel and point in the same direction, so that side $XP$ is parallel to side $QR.$ And also from $X=P+Q-R$ we get $X-Q=P-R$ showing the other pair $XQ,PR$ are parallel (the remaining pair of parallel sides of that parallelogram). Suggest draw a picture to see why for this case each of $XR$ and $PQ$ wind up as "diagonals" of the parallelogram formed.
Here, Three points are given,

$P(-2,\;2)$

$Q(1,\;6)$

$R(4,\;3)$

and take $S(x,\;y)$

Since in parallelogram

$\vec{PS}=\vec{QR}$,

Therefore,
$$(Sx−Px,Sy−Py)=(Rx−Qx,Ry−Qy)$$
$$⇒(x−(−2),y−2)=(4−1,3−6)$$
$$⇒x−(−2)=4-1\;\;\implies x=0$$ 
 $$⇒y−2=3−6\;\;\implies y=-1$$

Thus, coordinates of fourth vertex are $S(0,\;-1)$

I hope it' ll help.