# How do you solve n^2+2n-24>0?

Apr 21, 2017

$- 4 < n > 6$

#### Explanation:

${n}^{2} + \textcolor{red}{2} n \textcolor{\mathmr{and} a n \ge}{- 24} > 0$

We need to factor this quadratic

We need to find two numbers that add to $2$ and multiply to $- 24$
$+ 2$
x$24$
..........
$\textcolor{b l a c k}{1} \textcolor{w h i t e}{.} \textcolor{b l a c k}{24}$
$\textcolor{b l a c k}{2} \textcolor{w h i t e}{.} \textcolor{b l a c k}{12}$
$\textcolor{b l a c k}{3} \textcolor{w h i t e}{.} \textcolor{b l a c k}{8}$
$\textcolor{g r e e n}{4} \textcolor{w h i t e}{.} \textcolor{g r e e n}{6}$

$4$ and $6$ can be combined to form $2$, but we'll need to subtract them.

We are looking for one positive number and one negative number. We can tell because $- 24$ can only be the product of one positive and negative number.

Now we just need to see if $- 4 + 6$ gives us $\textcolor{red}{2}$ or if it's $- 6 + 4$:
$- 4 + 6 = 2$
$- 6 + 4 = - 2$

So, our numbers are $- 4$ and $6$, which gives us $\left(n - 4\right) \left(n + 6\right) > 0$

We just set each factor to zero and solve:
Case 1
$n - 4 > 0$

$n > 4$

Case 2
$n + 6 > 0$

$n > 6$