Katex

x2+1 \sqrt{x^2+1} x2+1​ ∑_i=1∞1n2=π26 \sum\_{i=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6} ∑_i=1∞n21​=6π2​ When a≠0 a \ne 0 a=0 , there are two solutions to (ax2+bx+c=0)(ax^2 + bx + c = 0)(ax2+bx+c=0) and they are x=−b±b2−4ac2a x = {-b \pm \sqrt{b^2-4ac} \over 2a} x=2a−b±b2−4ac​​