Test

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

$a \ne 0$
$ax^2 + bx + c = 0$
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

$$E_{xc}^{LDA}[n] = \int f_{xc} (n) d\bold{r} = \int n(\bold{r}) \epsilon_{xc}^{uniform} (n(\bold{r})) d\bold{r}$$

Vectors in $d^2$ have a shape of $dsa_dsa_ddds dsa_{dds} das_dds_dd$

$$\underbrace{a+b+c}_{\text{note dd}}$$ dsa

$$\underbrace{E_{xc}^{LDA}[n]}_{\text{functional of } n(\bold{r})} = \int d\bold{r} n(\bold{r})$$

$$\underbrace{\epsilon_{xc}^{unif}(n(\bold{r}))}_{\text{function of n at }\bold{r}}$$

$$\underbrace{E_{xc}^{LDA}[n]} _ {\text{functional of } n(\bold{r})} = \int d\bold{r} n(\bold{r}) \underbrace{\epsilon_{xc}^{unif}(n(\bold{r}))}_{\text{function of n at }\bold{r}}$$

$$\underbrace{E_{xc}^{LDA}[n]}{\text{functional of } n(\bold{r})} = \int d\bold{r} n(\bold{r}) \underbrace{\epsilon^{unif}(n(\bold{r}))}_{\text{function of n at }\bold{r}}$$