# How do you solve 21n ^ { 2} + 5n - 6= 0?

Mar 28, 2018

$n = \frac{3}{7} , - \frac{2}{3}$

#### Explanation:

Given:
$\textcolor{w h i t e}{\times x} 21 {n}^{2} + 5 n - 6 = 0$

$\Rightarrow 21 {n}^{2} - 9 n + 14 n - 6 = 0$ [Broke up $5 n$ as $- 9 n + 14 n$]

$\Rightarrow 3 n \left(7 n - 3\right) + 2 \left(7 n - 3\right) = 0$ [Grouping the Like Terms]

$\Rightarrow \left(7 n - 3\right) \left(3 n + 2\right) = 0$ [Group Again]

Now, If :

$3 n + 2 = 0$, then $3 n = - 2 \Rightarrow n = - \frac{2}{3}$

And If :

$7 n - 3 = 0$, then $7 n = 3 \Rightarrow n = \frac{3}{7}$

Hence Explained.

Mar 28, 2018

$n = - \frac{2}{3} \text{ or } n = \frac{3}{7}$

#### Explanation:

$\text{factorise the quadratic using the a-c method}$

$\text{that is factors of "21xx-6=-126" which sum to } + 5$

$\text{the factors of "-126" which sum to + 5 are + 14 and - 9}$

$\textcolor{b l u e}{\text{split the middle term using these factors}}$

$\Rightarrow 21 {n}^{2} + 14 n - 9 n - 6 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\Rightarrow \textcolor{red}{7 n} \left(3 n + 2\right) \textcolor{red}{- 3} \left(3 n + 2\right) = 0$

$\text{take out the common factor } \left(3 n + 2\right)$

$\Rightarrow \left(3 n + 2\right) \left(\textcolor{red}{7 n - 3}\right) = 0$

$\text{equate each factor to zero and solve for n}$

$3 n + 2 = 0 \Rightarrow n = - \frac{2}{3}$

$7 n - 3 = 0 \Rightarrow n = \frac{3}{7}$