How do you solve #x^2 - 3x - 10 = 0 #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer George C. Jun 14, 2015 #x^2-3x-10 = (x-5)(x+2)# So #x^2-3x-10=0# has roots #x=5# and #x=-2#. Explanation: If #x^2-3x-10 = (x-a)(x+b) = x^2-(a-b)x-ab# then #a*b = 10# and #a-b=3#. That is satisfied by #a=5# and #b=2#. 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 18246 views around the world You can reuse this answer Creative Commons License