How do you solve #3x² + 5x = 2#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer ali ergin Mar 12, 2016 #x_1=-2# #x_2=1/3# Explanation: #3x^2+5x=2# #3x^2+5x-2=0# #ax^2+bx+c=0# #Delta=sqrt(b^2-4*a*c)# #Delta=sqrt(5^2+4*3*2)# #Delta=sqrt(25+24)" "Delta =sqrt49 " "Delta=±7# #x_"1,2"=(-b±Delta)/(2*a)# #x_1=(-b-Delta)/(2*a)=(-5-7)/(2*3)=(-12)/6# #x_1=-2# #x_2=(-b+Delta)/(2*a)=(-5+7)/(2*3)=2/6# #x_2=1/3# 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 2118 views around the world You can reuse this answer Creative Commons License