Testing MathJax

When $a \ne 0$, there are two solutions to (ax^2 + bx + c = 0) and they are: $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

\[ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \]

This should be inline: \( ax^2 + \sqrt{bx} + c = 0 \)

Hi `$z = x + y$`

.

`$$a^2 + b^2 = c^2$$`

```
$$\begin{vmatrix}a & b\\
c & d
\end{vmatrix}=ad-bc$$
```

e have `\(x_1 = 132\)`

and `\(x_2 = 370\)`

and so ...

\begin{array}{cc} a & b \\ c & c \end{array}

When (a \ne 0), there are two solutions to (ax^2 + bx + c = 0) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

When (a \ne 0), there are two solutions to (ax^2 + bx + c = 0) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

... when $x < y$ we have ...