How do you solve #r^2+(sqrt(3)-sqrt(2))r = sqrt(6)#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Cesareo R. Aug 18, 2016 #r = sqrt(2)# and #r = -sqrt(3)# Explanation: #r^2+(sqrt(3)-sqrt(2))r = sqrt(6)->r^2+(sqrt(3)-sqrt(2))r-sqrt(2)sqrt(3)=0# then #(r-sqrt(2))(r+sqrt(3))=0# and the solutions are #r = sqrt(2)# and #r = -sqrt(3)# 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 1313 views around the world You can reuse this answer Creative Commons License