What does this equation mean?

Fix the mass number A and let Z vary. Because the Coulomb and asymmetry terms in the semi-empirical mass formula are both approximately quadratic in Z at fixed A , the atomic mass along an isobaric chain is, to good approximation, a parabola in Z : $M(A,Z)\,c^2 \approx \alpha(A) - \beta(A)\,Z + \gamma(A)\,Z^2 \mp \delta(A)$. where $\alpha$ , $\beta$ , and $\gamma$ collect the volume, surface, Coulomb, and asymmetry contributions and the pairing term $\delta(A)$ shifts the curve up or down depending on whether N and Z are both even or both odd. For odd A , exactly one of N , Z is even and the pairing term vanishes, giving a single parabola with one minimum: one value of Z at that mass number is the most tightly bound, and every other isobar decays toward it by beta-minus emission (if it sits to the neutron-rich side) or by positron emission or electron capture (if it sits to the proton-rich side). For even A , the pairing term splits the curve into two parabolas — one for even-even nuclei sitting lower in mass (more bound), one for odd-odd nuclei sitting higher — offset from each other by roughly 2\,$a_P$ $A^{-1/2}$ . This is the double-parabola structure behind one of the chart’s more striking features: a modest number of even mass numbers host two, and even three, beta-stable even-even isobars side by side, because an odd-odd nucleus between them would have to climb over the even-even parabola’s higher wall to decay by a single beta step in either direction, and single beta decay cannot skip that wall in one move. NUBASE2020’s compiled ground-state properties record this pattern directly, in the small set of mass numbers with more than one nuclide, of the same parity family, that show no observed beta decay at all [ 2 ] .