How do you factor and solve #x^2+5x-2=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Dee Apr 17, 2017 #x = -5/2 + sqrt(33)/2# and #x = -5/2 -sqrt(33)/2# Explanation: Use quadratic formula: #x = (-b + sqrt(b^2-4*a*c))/{2*a}# and #x = (-b - sqrt(b^2-4*a*c))/{2*a};# where #a = 1, b = 5; c = -2.# 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 14383 views around the world You can reuse this answer Creative Commons License