How do you find the solution to the quadratic equation #2x^(2/3) + 5^(1/3) = 12#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Massimiliano May 9, 2015 In this way: let #x^(1/3)=t# so #x^(2/3)=t^2#. #2t^2+5t-12=0# #Delta=b^2-4ac=25+4*2*12=121# #t=(-b+-sqrtDelta)/(2a)=(-5+-11)/4rArr# #t_1=(-5-11)/4=(-16)/4=-4# #t_2=(-5+11)/4=6/4=3/2#. And now: #x^(1/3)=-4rArr(x^(1/3))^3=(-4)^3rArrx=-64# #x^(1/3)=3/2rArr(x^(1/3))^3=(3/2)^3rArrx=27/8#. 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 1525 views around the world You can reuse this answer Creative Commons License