Question

How do you find the zeros, real and imaginary, of `y=3x^2-17x-9` using the quadratic formula?

Answer 1

`x_1 = (17 - sqrt397)/6` and `x_2 = (17 + sqrt397)/6`

Explanation:

You first need to calculate `b^2 - 4ac = Delta`. Here, `Delta = 289 + 4*3*9 = 289 + 108 = 397 > 0` so it has 2 real roots.

The quadratic formula tells us that the roots are given by `(-b +- sqrtDelta)/(2a)`.

`x_1 = (17 - sqrt397)/6` and `x_2 = (17 + sqrt397)/6`