How do you solve #5x^2-40x+80=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Kassi J May 12, 2016 I think the answer is -5.3. Explanation: #5x^2-40x+80 =0# #5x*5x=25x# Take care of the exponent first. #25x-40x+80=0# Solve the like terms. #-15x+80=0# Carry 80 over. #-15x/-15=80/-15# Cancel out -15 and divide 80 by that same number. #x=-5.3# Plug in the answer you get and round it to the nearest ten. 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 2137 views around the world You can reuse this answer Creative Commons License