How do you factor and solve #x^2+4=5x#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Konstantinos Michailidis May 18, 2016 We have that #x^2+4=5x=>x^2-x+4-4x=0=>x*(x-1)-4(x-1)=0=> (x-1)*(x-4)=0# Hence the solutions are #x_1=1# and #x_2=4# 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 2748 views around the world You can reuse this answer Creative Commons License