氺菓
氺菓
Nekor
氺菓
Nekor
氺菓
Nekor
氺菓
氺菓回复给帖子:1573720
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When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
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氺菓
氺菓回复给帖子:1573720
展开Biu
\begin{align}
a_{11}& =b_{11}& a_{12}& =b_{12}\notag\\
a_{21}& =b_{21}& a_{22}& =b_{22}+c_{22} \tag{y}
\end{align}
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