The proton (positive charge) shall be closer to the charge with the lower magnitude which is at the orgin.
The proton also shall be out of the interval between the two charges so that the pull of one charge cancels with the push of the other.
The region at which those conditions happen is to the left of the origin.
In that case the forces over the proton shall be:
k* (2.4 nC) * p / (x^2) - k*(4.8 nC) * p /( 1.3 + x)^2 = 0
where p is the charge of the proton.
You can simplify k and p:
2.4 / x^2 - 4.8 / (1.3 + x)^2 = 0
You can also simplify by 2.4
1/ x^2 - 2 / (1.3 + x)^2 = 0
(1.3+x)^2 - 2x^2 = 0
1.69 + 2.6x + x^2 - 2x^2 = 0
1.69 + 2.6x - x^2 = 0
x^2 -2.6x - 1.69 =0
Solve using the quadratic formula: x = 3.14 (use only the positive value)
That is the proton shall be place 3.14 units to the left of the origin (positive charge)
