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)#