How do you solve #7x^2+10x=2x^2+155# using any method? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Aug 12, 2016 #x=4sqrt2-1# or #x=4sqrt2+1# Explanation: #7x^2+10x=2x^2+155# #hArr7x^2+10x-2x^2-155=0# or #5x^2+10x-155=0# and dividing by #5# #x^2+2x-31=0# and using quadratic formula #(-b+-sqrt(b^2-4ac))/2a# , as we have here #a=1#, #b=2# and #c=-31# #x=(-2+-sqrt((-2)^2-4×1×(-31)))/(2×1)# or #x=(-2+-sqrt(4+124))/(2×1)# or #x=(-2+-sqrt128)/2# or #x=(-2+-8sqrt2)/2=-1+-4sqrt2# or #x=4sqrt2-1# or #x=4sqrt2+1# 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 1281 views around the world You can reuse this answer Creative Commons License