How do you solve by factoring # 5n^2- 17n= -6#?

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Answer 1 · 2016-05-14T21:41:49

#n = 2/5# or #n = 3#

First add #6# to both sides to get:

#5n^2-17n+6 = 0#

Then use an AC method:

Look for a pair of factors of #AC = 5*6 = 30# with sum #B=17#.

The pair #15, 2# works.

Use this pair to split the middle term and factor by grouping:

#5n^2-17n+6#

#= 5n^2-15n-2n+6#

#= (5n^2-15n)-(2n-6)#

#= 5n(n-3)-2(n-3)#

#= (5n-2)(n-3)#

So our original equation becomes:

#(5n-2)(n-3) = 0#

with solutions:

#n = 2/5# or #n = 3#