How do you solve #2x²-(5/2)x+2=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Jul 2, 2017 #x=5/8+isqrt39/8# or #5/8-isqrt39/8# Explanation: #2x^2-5/2x+2=0# can be written as #2(x^2-5/4 x+1)=0# - dividing by #2# or #x^2-5/4 x+1=0# or #x^2-2xx5/8xx x+(5/8)^2-(5/8)^2+1=0# or #(x-5/8)^2-25/64+1=0# or #(x-5/8)^2+39/64=0# or #(x-5/8)^2-(-1xx39/64)=0# or #(x-5/8)^2-(isqrt39/8)^2=0# or #(x-5/8+isqrt39/8)(x-5/8-isqrt39/8)=0# Hence #x=5/8+isqrt39/8# or #5/8-isqrt39/8# 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 955 views around the world You can reuse this answer Creative Commons License