# How do you solve (n+2) (2n+5)= 0 by factoring?

Aug 17, 2015

The solutions are

 color(blue)(n=-2

 color(blue)(n=-5/2

#### Explanation:

$\left(n + 2\right) \left(2 n + 5\right) = 0$ is the already factorised form of equation
$2 {n}^{2} + 9 n + 10$.

Since we have the factors we obtain the solution by equating these factors to the R.HS.

n+2=0, color(blue)(n=-2
2n+5=0, color(blue)(n=-5/2