First Post

When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ A Cross Product Formula \[\mathbf{V}_1 \times \mathbf{V}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial X}{\partial u} & \frac{\partial Y}{\partial u} & 0 \\ \frac{\partial X}{\partial v} & \frac{\partial Y}{\partial v} & 0 \end{vmatrix} \]

Math in TeX notation

When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ $$ \begin{array}{rcll} y & = & x^{2}+bx+c\\ & = & x^{2}+2\times\dfrac{b}{2}x+c\\ & = & \underbrace{x^{2}+2\times\dfrac{b}{2}x+\left(\frac{b}{2}\right)^{2}}- {\left(\dfrac{b}{2}\right)^{2}+c}\\ & & \qquad\left(x+{\dfrac{b}{2}}\right)^{2}\\ & = & \left(x+\dfrac{b}{2}\right)^{2}-\left(\dfrac{b}{2}\right)^{2}+c & \left|+\left({\dfrac{b}{2}}\right)^{2}-c\right.\\ y+\left(\dfrac{b}{2}\right)^{2}-c & = & \left(x+ \dfrac{b}{2}\right)^{2} & \left|\strut(\textrm{vertex form})\right.\\ y-y_{S} & = & (x-x_{S})^{2}\\ S(x_{S};y_{S}) & \,\textrm{or}\, & S\left(-\dfrac{b}{2};\,\left(\dfrac{b}{2}\right)^{2}-c\right) \end{array} $$ Experimental Post