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 ...