hello world!
january 1, 2024
just testing the blog page, it works apparently :)
Math in TeX notation
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
$$ \begin{array}{rcll}
y & = & x^{2}+bx+c\\
& = & x^{2}+2\times\dfrac{b}{2}x+c\\
& = & \underbrace{x^{2}+2\times\dfrac{b}{2}x+\left(\frac{b}{2}\right)^{2}}-
{\left(\dfrac{b}{2}\right)^{2}+c}\\
& & \qquad\left(x+{\dfrac{b}{2}}\right)^{2}\\
& = & \left(x+\dfrac{b}{2}\right)^{2}-\left(\dfrac{b}{2}\right)^{2}+c
& \left|+\left({\dfrac{b}{2}}\right)^{2}-c\right.\\
y+\left(\dfrac{b}{2}\right)^{2}-c & = & \left(x+
\dfrac{b}{2}\right)^{2} & \left|\strut(\textrm{vertex form})\right.\\
y-y_{S} & = & (x-x_{S})^{2}\\
S(x_{S};y_{S}) & \,\textrm{or}\,
& S\left(-\dfrac{b}{2};\,\left(\dfrac{b}{2}\right)^{2}-c\right)
\end{array} $$