Solving inequalities. How to solve `(x+5)/(3-x^2)≥ 0`?

A fraction is positive or zero if and only if numerator and denominator have the same sign

Case 1.- Both positives
`x+5>=0` then `x>=-5` and
`3-x^2>0` (imposible to be zero) then `3>x^2` that is

`-sqrt3 < x < sqrt3`

The intersection of both sets of values is `[-5,oo)nn(-sqrt3,sqrt3)=(-sqrt3,sqrt3)`

Case 2.- Both negatives

Similarly the solutions are `(-oo,-5]nn((-oo,-sqrt3)uu(sqrt3,+oo))=`

`=[-5,-sqrt3)uu(sqrt3,+oo)`

Now, the union of both cases will be the final result

`[-5,-sqrt3)uu(-sqrt3,sqrt3)uu(sqrt3,+oo)`

The inequality is

`(x+5)/(3-x^2)>=0`

`(x+5)/((sqrt3-x)(sqrt3+x))>=0`

Let `f(x)=(x+5)/((sqrt3-x)(sqrt3+x))`

Let's build the sign chart

`color(white)(aaaa)``x``color(white)(aaaa)``-oo``color(white)(aaaa)``-5``color(white)(aaaa)``-sqrt3``color(white)(aaaa)``+sqrt3``color(white)(aaaa)``+oo`

`color(white)(aaaa)``x+5``color(white)(aaaa)``-``color(white)(aaa)``0``color(white)(aaa)``+``color(white)(aaaaa)``+``color(white)(aaaaa)``+`

`color(white)(aaaa)``sqrt3+x``color(white)(aaa)``-``color(white)(aaa)`#color(white)(aaa)-#`color(white)(aaa)``||``color(white)(aa)``+``color(white)(aaaaa)``+`

`color(white)(aaaa)``sqrt3-x``color(white)(aaa)``+``color(white)(aaa)`#color(white)(aaa)+#`color(white)(aaa)`#color(white)(aaa)+#`color(white)(aa)``||``color(white)(aa)``-`

`color(white)(aaaa)``f(x)``color(white)(aaaaaa)``+``color(white)(aaa)``0``color(white)(aa)``-``color(white)(aaa)``||``color(white)(aa)``+``color(white)(aa)``||``color(white)(aa)``-`

Therefore,

`f(x)>=0` when `x in (-oo,-5]uu(-sqrt3, sqrt3)`

graph{(x+5)/(3-x^2) [-12.66, 12.66, -6.33, 6.33]}