What is the solution set for # x^2 + 6x + 10 = 0#? Algebra Quadratic Equations and Functions Quadratic Formula 1 Answer Nallasivam V Aug 10, 2015 #x =(-6+ sqrt( - 4))/ (2)# #x =(-6- sqrt( - 4))/ (2)# Explanation: Since ------- #6^2# - #(4 xx1 xx 10)# < 0, x has imaginary roots #x =(-b+- sqrt(b^2 - (4ac)))/ (2a)# #x =(-6+- sqrt(6^2 - (4xx1xx10)))/ (2xx1)# #x =(-6+- sqrt(36 - 40))/ (2)# #x =(-6+ sqrt( - 4))/ (2)# #x =(-6- sqrt( - 4))/ (2)# Answer link Related questions How do you know how many solutions #2x^2+5x-7=0# has? What is the Quadratic Formula? How do you derive the quadratic formula? How is quadratic formula used in everyday life? How do you simplify the quadratic formula? How do you solve #x^2+10x+9=0# using the quadratic formula? How do you solve #-4x^2+x+1=0# using the quadratic formula? When do you have "no solution" when solving quadratic equations using the quadratic formula? How do you solve #4x^2=0# using the quadratic formula? Why can every quadratic equation be solved by using the quadratic formula? See all questions in Quadratic Formula Impact of this question 4385 views around the world You can reuse this answer Creative Commons License