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Art of Problem Solving
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and language arts learners in grades 2-12.
(that logo typeset by ![\[\text{If }ax^2+bx+c=0\text{, then }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]](http://latex.artofproblemsolving.com/7/c/b/7cbf48dbdce18ee42c347668a9314410de651207.png)
![\[e^{i\pi}+1=0\]](http://latex.artofproblemsolving.com/2/6/c/26cc6b1c790a7db2144e4f5948158479292c55e6.png)
![\[\sum_{x=1}^{\infty} \frac{1}{x}=2\]](http://latex.artofproblemsolving.com/e/e/5/ee5784ec0e67d8ddb8d721443e25e5a727e1cf76.png)
![\begin{align*} x &= \sqrt[3]{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right) + \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} - \frac{b^2}{9a^2}\right)^3}} \\ &+ \sqrt[3]{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right) - \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} - \frac{b^2}{9a^2}\right)^3}} - \frac{b}{3a} \\ &\text{(I copied it from another website but I typeset it myself; I am pretty sure those are not copyrightable. I still need \textit{years} to understand this.)} \end{align*}](http://latex.artofproblemsolving.com/6/4/a/64a63f941965a9c236d28753106dfac10ca4b515.png)
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