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16n^2-4=0
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ok first of all welcome to open study \[\large{(4n)^2-(2)^2=(4n+2)(4n-2)}\]
do you want to find the value of n or just factor this ?
the value of n
16n^2=4 n^2=1/4 n=+- 1/2
\[\large{(4n+2)(4n-2)=0}\] \[\large{4n=-2 , 4n=2}\] \[\large{n=-1/2 , 1/2}\]
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