Solve the inequality on the number line? How would I solve the following inequality.
$x^2+10x \gt-24$
How would I solve it and put it in a number line?
 A: Hint: solve for 
$x^2+10x+ 24 = 0$, find the roots, and determine when the factors are positive: 
Solve: $$x^2 + 10 x + 24 = (x+4)(x+6) > 0$$
When is $(x+4)(x + 6)$ positive?:
$\quad$When both factors are positive, or when both factors are negative.  
$$
(x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0, & (x+6) > 0 \longrightarrow x>-4\\ \\(x+4) < 0, & (x+6) < 0 \longrightarrow x<-6 \end{cases}
$$
Your task is to plot the intervals on which $x$ satisfies the inequality.

Edit: if you want to confirm the solution "graph", compare to:

A: $$x^2 + 10x  > -24 \implies x^2 + 10x + 24 > 0 \implies (x+4)(x+6) > 0$$
Recall that if $ab > 0$, then either $a>0 \,\, \& \,\, b > 0$ or $a < 0 \,\, \& \,\, b < 0$.
Hence, $$(x+4)(x+6) > 0 \implies \begin{cases}(x+4) > 0; & (x+6) > 0 \implies x>-4\\ (x+4) < 0; & (x+6) < 0 \implies x<-6 \end{cases}$$
Hence, we get that $$x \in (-\infty,-6) \cup (-4, \infty)$$
A: Solve the quadratic $x^2+10x+24=0$ and see what happens between the two roots.
A: $x^2+10x>−24$
iff $x^2-10x+25 > 1$
iff $(x-5)^2 > 1$.
(Note: "iff" means "if and only if".)
Because of the magical property of $1$ being its own square root
(funny how this happens, eh?)
this is true iff $|x-5| > 1$
which is true
iff $x<4$ or $x > 6$.
By completing the square like this,
you are implicitly solving the equation.
If you start with the inequality
$x^2-2bx > c$
($b=5$ and $c=-24$ in your case)
this becomes
$(x-b)^2 > c + b^2$.
If $c + b^2 < 0$,
all $x$ satisfy this;
otherwise this can be rewritten as
$|x-b| > \sqrt{c+b^2}$
for which the solutions are
$x < b-\sqrt{c+b^2}$
and
$x > b+\sqrt{c+b^2}$
If you start with the inequality
$x^2-2bx < c$
this becomes
$(x-b)^2 < c + b^2$.
If $c + b^2 < 0$,
no $x$ satisfies this;
otherwise this can be rewritten as
$|x-b| < \sqrt{c+b^2}$
for which the solutions are
$b-\sqrt{c+b^2}
 < x < b+\sqrt{c+b^2}$
