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Formulae used in JEE

The two solutions to \(ax^2 + bx + c = 0\) are, $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ when $a \ne 0$



Sum of Squares of first Naturals Numbers is :- $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$