Binary
About Me
Interests
Timetable
Contact
Interests
My Hobbies
Swimming
Guitar
Squash
Future Goals
Take part in ICPC
A Foreign Internship
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 :- $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$