The potential is negative in the region surrounding the central, negative charge, and is positive surrounding each of the positive charges. At a great distance, the potential is positive and approaches the value for a single positive point charge $q.$ Thus an equipotential surface on which the potential is zero surrounds the negative charge, as shown in the diagram.
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On the $x-$axis, with $x<d,$ the potential is
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and so we may find the point where $V=0:$
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or
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Solving the quadratic, we get:
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Since we started out assuming $0<x<d,$ we must choose the plus sign, and then
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