How do you solve #2x^2-8x=-8#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. May 17, 2016 Answer: #x=2# Explanation: If #color(white)("XXX")2x^3-8x=-8# then #color(white)("XXX")x^2-4x=-4# (after dividing by #2#) #color(white)("XXX")x^2-4x+4=0# (after adding #4# to both sides) #color(white)("XXX")(x-2)^2=0# (factoring) #color(white)("XXX")(x-2)=0# (since if #a*b=0# then #a=0# or #b=0#) #color(white)("XXX")x=2# (after adding #2# to both sides) 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 830 views around the world You can reuse this answer Creative Commons License