How do you find the zeros, real and imaginary, of #y=9x^2-4x-15# using the quadratic formula?

1 Answer
Jul 30, 2016

Answer:

1.53 and - 1.09

Explanation:

#y = 9x^2 - 4x - 15 = 0#
Use the new quadratic formula in graphic form (Socratic Search).
#D = d^2 = b^2 - 4ac = 16 + 540 = 556# --> #d = +- 23.58#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 4/18 +- 23.58/18 = (4 +- 23.58)/18#
#x1 = 27.58/18 = 1.53#
#x2 = - 19.58/18 = - 1. 09#
graph{9x^2 - 4x - 15 [-40, 40, -20, 20]}