How do you solve #2x^2-13x-7=0#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer BRIAN M. Jun 4, 2016 #x = -1/2 or x= 7# Explanation: The trinomial #2x^2-13x-7=0# can be factored #(2x +1)(x-7)=0# Now set each factor equal to zero #2x+1=0# #2xcancel(+1)cancel(-1)=0-1# #2x=-1# #(cancel(2)x)/cancel2=-1/2# #x=-1/2# and #x-7=0# #xcancel(-7)cancel(+7)= 0+7# #x = 7# 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 2300 views around the world You can reuse this answer Creative Commons License