Question

find the number of real-number solutions of 4x^2+16x+15=0

Answer 1

Using Descartes' Rule of Signs, you see how many times the sign changes between the terms. There are 0 positive roots. Then, plug in (-x) for every x and see how many times the sign changes. 4x^2 - 16x + 15 = 0. This means that there can be 2 or 0 negative roots. However in this case, there are 2 real negative solutions.