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

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

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Answer 1 · 2015-05-14T11:51:16

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

Temporarily extract the constant from the working left side

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

Extract the $m (= 5)$ $m (=5)$ factor

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

Complete the square

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

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

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

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

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

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