How do you solve #x ^2 + 2x - 22 = 0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer G_Ozdilek Apr 4, 2017 3.796 and -5.796 Explanation: Compute #(b^2)-(4*a*c)# first which is delta. b is 2 and c is -22 (a is 1 in this equation). You will get delta as 92. #x=(-b+sqrt92)/2=3.796# the first x #x=(-b-sqrt92)/2=-5.796# the second x 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 2104 views around the world You can reuse this answer Creative Commons License