Determine the solubility of HgBr2, a sparingly soluble salt, in the presence of 0.0010 M NH3.  NH3 is a weak base that reacts with Hg2+ in the reaction Hg2+ + 4NH3 (aq) ↔ Hg(NH3)42+. Using the Systematic Method.

 Hg2+ + 4NH3 (aq) ↔ Hg(NH3)42+
ammonium complexes are completely soluble.
HgBr2 (s)<---->Hg2+  + Br-
Ksp = [Hg2+][2Br-]^2
if the solubility is x then amount of  Hg2+ will be x + 0.0010 and [Br-] is x
6.2*10^-20 =(x+0.0010)(2x+0.0020)^2
solve it for x