Suppose third charge is similar to Q and it is q

So net force on it

F net = 2F cos θ θ

Where $F = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q q}{\left( x^2 + \frac{d^2}{4} \right)^{3/2}}$ and $\cos \theta = \frac{x}{\sqrt{x^{2} + \frac{d^{2}}{4}}}$

∴ ∴ $F_{net} = 2 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Qq}{\left(x^2 + \frac{d^2}{4}\right)} \times \frac{x}{\left(x^2 + \frac{d^2}{4}\right)^{1/2}}$

$= \frac{2Qqx}{4 \pi \varepsilon_0 \left( x^2 + \frac{d^2}{4} \right)^{3/2}}$

For F net to be maximum $\frac{dF_{\text{net}}}{dx} = 0$

i.e. $\frac{d}{dx}\left[\frac{2Qqx}{4\pi\epsilon_0\left(x^2+\frac{d^2}{4}\right)^{3/2}}\right]=0$

or $\left[\left(x^{2}+\frac{d^{2}}{4}\right)^{-3/2} - 3x^{2}\left(x^{2}+\frac{d^{2}}{4}\right)^{-5/2}\right] = 0$

i.e. $x = \pm \frac{d}{2 \sqrt{2}}$