How do you solve #x^2 + x = 3/4#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer P dilip_k Mar 23, 2016 #=>4x^2+4x-3=0# #=>4x^2+6x-2x-3=0# #=>2x(2x+3)-1(2x+3)=0# #=>(2x+3)(2x-1)=0# #:. x=-3/2 and x=1/2# 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 1025 views around the world You can reuse this answer Creative Commons License