# How do you solve 14n^2 + 24n - 5004 = 0?

Mar 5, 2018

#### Explanation:

The straightforward way is to use the quadratic formula:
$n = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
This is for expressions of the form $a {n}^{2} + b n + c$, so in your case a=14, b=24, c=-5004 .

A slightly more complicated way is to reverse foil, trying to factor out your a, b, and c so you can write you equation into
$\left(x n + y\right) \left(s n + t\right) = 0$, where x, y, s, and t are all numbers. Then you can solve one $\left(\right)$ at a time. But if you have a calculator, the quadratic formula is the best way to go. It'll work every time.

Mar 5, 2018

$n = - \frac{6}{7} \pm \frac{15}{7} \sqrt{78}$

#### Explanation:

Complete the square then use the difference of squares identity to find:

$0 = \frac{7}{2} \left(14 {n}^{2} + 24 n - 5004\right)$

$\textcolor{w h i t e}{0} = 49 {n}^{2} + 84 n - 17514$

$\textcolor{w h i t e}{0} = {\left(7 n\right)}^{2} + 2 \left(7 n\right) \left(6\right) + 36 - 17550$

$\textcolor{w h i t e}{0} = {\left(7 n\right)}^{2} + 2 \left(7 n\right) \left(6\right) + {6}^{2} - \left({15}^{2} \cdot 78\right)$

$\textcolor{w h i t e}{0} = {\left(7 n + 6\right)}^{2} - {\left(15 \sqrt{78}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(7 n + 6\right) - 15 \sqrt{78}\right) \left(\left(7 n + 6\right) + 15 \sqrt{78}\right)$

$\textcolor{w h i t e}{0} = \left(7 n + 6 - 15 \sqrt{78}\right) \left(7 n + 6 + 15 \sqrt{78}\right)$

Hence:

$7 n = - 6 \pm 15 \sqrt{78}$

So:

$n = - \frac{6}{7} \pm \frac{15}{7} \sqrt{78}$