Open silverwind opened 2 years ago
When $a \ne 0$, there are two solutions to $(ax^2 + bx + c = 0)$ and they are $$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$
Does it need $\sqrt{3}$ ?
$$
x = {-b \pm \sqrt{b^2-4ac} \over 2a}
$$
\[
x = {-b \pm \sqrt{b^2-4ac} \over 2a}
\]
yup looks like fenced blocks don't work
$x$ -$x$ $x$-
a$xa$ $xa$a 1$xb$ $xb$1
$a a$b b$
a$b $a a$b b$
$a a\$b b$
Here is some math!