Diverse Ion effect and Activity coefficients problem

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The discussion revolves around calculating the solubility of Ba(IO3)2 in a 0.100M KIO3 solution at 25°C using activity coefficients. The Ksp value for Ba(IO3)2 is given as 1.57E-9, and the ionic strengths and activity coefficients for Ba2+ and IO3- are determined. The user sets up an ICE table to express the concentrations but becomes confused about solving for X due to the cubic nature of the equation. The main challenge lies in simplifying the equation to find a single solution for X, as multiple solutions complicate the calculation. Clarification on handling the cubic equation and the impact of activity coefficients is sought.
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The question:

How many grams of Ba(IO3)2 can be dissolved in 700mL of a 0.100M KIO3 solution at 25C? use activity coefficients for this. (Ksp of Ba(IO3)2 =1.57E-9)

my answer:

I used this equation: Ksp = (concentration) * (activity coefficient)

well i found out that my ionic strength of KIO3 is 0.100M and from a chart, i found the respective activity coeff for Ba2+ and IO3-.
Ba2+ ---> 0.38
IO3^- ----> 0.775

Ba(IO3)2 <----> Ba2+ + 2IO3^-
I 0 x 0.100M
C +x +2x
E x 0.100M+2x


using the equation above,
Ksp = [Ba2+]*(activity coeff of Ba) * [IO3^-]^2 * (activity coeff of IO3^-)^2
1.57E-9 = (x)(0.38) * (0.100+2x)^2 * (0.775)^2

so i solve for X and that would be the concentration of Ba(IO3)2 in mol/L

BUT, to solve for X, i found it complicated because there would be 3 solutions for X aren't there?? since it's X^3... so i am confused on what to do.

help appreciated, Thanks.
 
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there above numbers below the balanced equation is supposed to be an ICE table
 

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