How do you solve #-15x^2 + 18x = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Dean R. Apr 30, 2018 First factor: #0 = -15 x^2 + 18 x = -3 x(5x - 6)# so #x=0# or #5x-6=0#, that is, #x=6/5.# Explanation: Check: # -15(0)^2 + 18(0) = 0 quad sqrt# #-15(6/5)^2 + 18(6/5)# # = -15(36/25) + {3(36)}/5 # #= -{3(36)}/5 + {3(36)}/5 = 0 quad sqrt# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 2083 views around the world You can reuse this answer Creative Commons License