Question 1030456
Multiply,
{{{n^2+4n=96}}}
{{{n^2+4n-96=0}}}
{{{(n+12)(n-8)=0}}}
Two solutions:
{{{n+12=0}}}
{{{n=-12}}}
So then,
{{{n+4=-8}}}
-12 and -8
and
{{{n-8=0}}}
{{{n=8}}}
then,
{{{n+4=12}
8 and 12