# How do you write 5n^2+19n-68=-2 into vertex form?

May 14, 2015

$5 {n}^{2} + 19 n - 68 = - 2$
to be written in the form:
$m {\left(n - a\right)}^{2} + b = 0$

Temporarily extract the constant from the working left side

$5 {n}^{2} + 19 n = 66$

Extract the $m \left(= 5\right)$ factor

$5 \left({n}^{2} + \frac{19}{5} n\right) = 66$

Complete the square

$5 \left({n}^{2} + \frac{19}{5} n + {\left(\frac{19}{10}\right)}^{2}\right) = 66 + 5 {\left(\frac{19}{10}\right)}^{2}$

$5 {\left(n + \frac{19}{10}\right)}^{2} = 66 + \frac{361}{20} = \frac{1681}{20}$

Move the constant back to the left side to complete the vertex form:

$5 {\left(n + \frac{19}{10}\right)}^{2} - \frac{1681}{20} = 0$
or
$5 {\left(n - \left(- \frac{19}{10}\right)\right)}^{2} + \left(- \frac{1681}{20}\right) = 0$