Two ways: a) Using the quadratic formula: The solution to a x^2 + b x + c is (-b +/- sqrt(b^2 - 4 a c)) / (2 a) b) by "completing the square" To complete the square, first divide through by the coefficient of x^2 to leave you with x^2 at the front. x^2 - 16 x - 32 = 0 Two identities: (x + a)^2 = x^2 + 2 a + a^2 (x - a)^2 = x^2 - 2 a + a^2 So x^2 + 2 a = (x + a)^2 - a^2 x^2 - 2 a = (x - a)^2 - a^2 So if you let a=8 be half the coefficient 16 you can use one of these formulae to turn x^2 - 16x - 32 = 0 into (x - 8)^2 - 64 - 32 = 0 and it's a simple matter to turn that into: x = 8 +/- sqrt(64 + 32) If your trinomial begins with 'x^2' and has integer coefficients, like this one does, with a bit of practice method (b) is MUCH easier! EDIT: I'm not happy that my explanation of completing the square is simple enough, so I'll try explaining it another way: You have: x^2 - 16x - 32 = 0 Now x^2 - 2*8 x + 8*8 would be a perfect square. It is (x-8)^2 In x^2 - 16 x - 32 = 0, x^2 - 16 x is almost a perfect square. I can make x^2 - 16x into a perfect square by adding 8*8 to both sides. x^2 - 16 x + 8*8 - 32 = 8*8 (x-8)^2 - 32 = 64 x - 8 = sqrt(64 + 32)