How do you solve #(2x+1)^2=x+2#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Rhys Jun 25, 2018 Shown below: Explanation: Expand and rearrange: #4x^2 +4x + 1 = x+2 # #=> 4x^2 +3x -1 = 0 # Factor: #(4x-1)(x+1) = 0 # #4x -1 = 0 => x = 1/4 # #x+1 = 0 => x = -1 # #x = {-1,1/4 } # 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 1268 views around the world You can reuse this answer Creative Commons License