#x^3+5x^2-x>=5# is equivalent to #x^3+5x^2-x-5>=0#

#x^3+5x^2-x-5=x(x^2-1)+5(x^2-1)=(x+5)(x^2-1)=(x+5)(x+1)(x-1)#

Hence, we have #(x+5)(x+1)(x-1)>=0#

From this we know that the product #(x+5)(x+1)(x-1)>=0# is positive or zero. It is apparent that sign of binomials #(x+5)#, #(x+1)# and #(x-1)# will change around the values #-5#. #-1# and #1# respectively.

In sign chart we divide the real number line in four parts, i.e. below #-5#, between #-5# and #-1#, between #-1# and #1# and above #1# and see how the sign of #(x+5)(x-1)(x+1)# changes.

**Sign Chart**

#color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)-1color(white)(XXXXX)1#

#(x+5)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XXX)+ive color(white)(XXX)+ive#

#(x+1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XXX)+ive color(white)(XXX)+ive#

#(x-1)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XXX)-ive color(white)(XXX)+ive#

#(x+5)(x+1)(x-1)#

#color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#

It is observed that #(x+5)(x+1)(x-1)>=0# when either #-5 <= x <= -1# or #x >= 1#, which is the solution for the inequality.