How do you solve #x^2+3x+1=0# algebraically? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Binayaka C. Nov 7, 2016 #x=(-3+sqrt5)/2 orx=(-3-sqrt5)/2# Explanation: #x^2+3x+1=0 or x^2+3x+9/4 -9/4+1=0 or (x+3/2)^2 -5/4=0 or (x+3/2)^2 =5/4 or x+3/2 = +-sqrt5/2 or x= -3/2+-sqrt5/2 :. x=(-3+sqrt5)/2 orx=(-3-sqrt5)/2# {Ans] 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 1680 views around the world You can reuse this answer Creative Commons License