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#