How do you find the zeroes for #f(x) = x^2 + x - 3#? Algebra Polynomials and Factoring Zero Product Principle 1 Answer Nghi N. Jun 3, 2015 f(x) = x^2 + x - 3 = 0 #D = d^2 = b^2 - 4ac = 1 + 12 = 13 -> d = +-sqr13# #x = -b/(2a) +-d/(2a) = -1/2 +- (sqrt13)/2# Answer link Related questions What is the Zero Product Principle? How to use the zero product principle to find the value of x? How do you solve the polynomial #10x^3-5x^2=0#? Can you apply the zero product property in the problem #(x+6)+(3x-1)=0#? How do you solve the polynomial #24x^2-4x=0#? How do you use the zero product property to solve #(x-5)(2x+7)(3x-4)=0#? How do you factor and solve #b^2-\frac{5}{3b}=0#? Why does the zero product property work? How do you solve #(x - 12)(5x - 13) = 0#? How do you solve #(2u+7)(3u-1)=0#? See all questions in Zero Product Principle Impact of this question 1297 views around the world You can reuse this answer Creative Commons License