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Magister Artium Liberalium
See also: Master's degree, MBA, M.D., J.D., Doctor of Philosophy, Degree
... b/(2a) from both sides, we get <math>x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math> ...