How do you solve #2x^2 - 2x = 1#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Nghi N. May 31, 2015 #y = 2x^2 - 2x - 1 = 0# #D = d^2 = b^2 - 4ac = 4 + 8 = 12 --> d = 2sqrt3# #x = -b/(2a) +- d/(2a) = 1/2 +- (sqrt3)/2# 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 1030 views around the world You can reuse this answer Creative Commons License