A charge of -6.6 µC is located at the origin; a charge of 8.8 µC is located at x

Question

A charge of -6.6 µC is located at the origin; a charge of 8.8 µC is located at x = 0.2 m, y = 0; a third charge Q is located at x = 0.32 m, y = 0. The force on the 8.8 µC charge is 1161.6 N, directed in the positive x direction.

Determine the charge Q.

I got : Q= -211.2

With this configuration of three charges, where, along the x direction, is the electric field zero?
x1(lower)= ??? x2(higher)= ???
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I can't seem to set it up any other way that doesn't give me a 4th root polynomial and irrational roots. I'm not sure what i'm doing wrong.

I have :

0= (-211.2/d^2) + (8.8 /(d+.12)^2) - (6.6/(d+.32)^2)

Explanation / Answer

the charge Q needs to be a negative charge so that the direction of the net electric force on 8.8uC is directed towards +x direction.

So, net electric force on 8.8 uC :

F= k*(8.8u)*( Q/(0.32 - 0.2)^2 - 6.6u/0.2^2) = 1161.6

So, 9*10^9*(8.8*10^-6)*(Q/0.12^2 - 6.6*10^-6/0.2^2) = 1161.6

So, Q = 213.6 uC

Now, as Q needs to be -ve, So, Q = -213.6 uC <------answer

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