Let's rewrite the inequality

#x^2-4x-5<=0#

Let's factorise the LHS

#(x+1)(x-5)<=0#

Let #f(x)=(x+1)(x-5)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaaa)##5##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+1##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-5##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 [ -1, 5 ]#