How do you solve # -x^2 + 100x - 2400 >= 0#? Algebra Linear Inequalities and Absolute Value Multi-Step Inequalities 1 Answer De Rono Oct 27, 2015 # 40 le x le 60# Explanation: #-x^2+100x-2400 ge 0# #implies -(x^2 -100x +2400) ge 0# #implies x^2 -100x +2400 le 0# #implies x^2 -100x +2500 le 100# #implies (x -50)^2 le 10^2# #implies -10 le x-50 le 10# #implies 40 le x le 60# Answer link Related questions How do you solve multi step inequalities? What is the difference between solving multi step equations and multi step inequalities? How do you solve multi step inequalities with variables on both sides? How do you solve for x given #x-5>x+6 #? What do you do when your variable cancels out? How do you solve for x when you have #4x-2(3x-9) \le -4(2x-9)#? How do you graph #\frac{5x-1}{4} > -2 (x+5)#? How do you solve for x in #4-6x \le 2(2x+3)#? How do you solve -3x+4<-8#? Which of these values of x satisfies the inequality #-7x+6≤ -8#? See all questions in Multi-Step Inequalities Impact of this question 3025 views around the world You can reuse this answer Creative Commons License