How do you factor and solve #m^2=-6m-7#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer EZ as pi · Mouhamed Ayoub E. Mar 11, 2017 #m =sqrt2-3" or " m=-sqrt2-3# Explanation: #m^2 =-6m-7# #m²+6m+9=-7+9" add "(6/2)^2# to both sides #(m+3)²=2# #(m+3)^2 - 2=0" "# factor as difference of squares #(m+3+sqrt2) (m+3-sqrt2)=0# #m+3=sqrt2 " or " m+3=-sqrt2# #m = sqrt2-3" or "m= -sqrt2 -3# 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 2189 views around the world You can reuse this answer Creative Commons License