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 ]#