How do you solve for x in #a^2x+b^2x+7=3c#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Noah G Nov 8, 2016 Isolate all terms with #x# on one side of the equation. #a^2x + b^2x = 3c - 7# #x(a^2 + b^2) = 3c - 7# #x = (3c - 7)/(a^2 + b^2)# Hopefully this helps! 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 1668 views around the world You can reuse this answer Creative Commons License