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

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Answer 1 · 2015-05-14T11:51:16

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

Temporarily extract the constant from the working left side

$5n^2+19n = 66$

Extract the $m (=5)$ factor

$5(n^2+19/5n) = 66$

Complete the square

$5(n^2+19/5n+(19/10)^2) = 66 + 5(19/10)^2$

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

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

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

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