Let's factorise the inequality
#x^2+7x+10=(x+2)(x+5)#
So,
#(x+2)(x+5)>0#
Let #f(x)=(x+2)(x+5)#
We build a sign chart
#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##-2##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x+5##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x+2##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)>0# when # x in (-oo,-5) uu (-2,+oo)#