Let's rewrite the inequality and factorise
#x^2-6x+9-16>0#
#x^2-6x-7>0#
#(x-7)(x+1)>0#
Let #f(x)=(x-7)(x+1)#
We build the sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##7##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-7##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)>0#, when #x in ]-oo,-1[uu]7,+oo[#