How do you solve #m^2+a+-16=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Alan P. Apr 7, 2016 #m=+-sqrt(-16-a)# (two solutions if #a<=-16#; none otherwise) or #m=+-sqrt(16-a)# (two solutions if #a<16#; none otherwise) Explanation: #m^2+a+-16=0# #rarr m^2= +-16-a# #rarr m=+-sqrt(+-16-a)# 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 1114 views around the world You can reuse this answer Creative Commons License