How do you solve #6x=-x²-8#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Suren Abreu Jan 24, 2017 #x=-2# or #x=-4# Explanation: #6x=-x^2-8# Add #x^2+8# to both sides. #x^2+6x+8=0# Factorise. #x^2+4x+2x+8=0# #x(x+4)+2(x+4)=0# #(x+2)(x+4)=0# #x+2=0# or #x+4=0# #x=-2# or #x=-4# 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 1067 views around the world You can reuse this answer Creative Commons License