How do you find the zeros, real and imaginary, of $y= -7x^2-3x-16$ using the quadratic formula?

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Answer 1 · 2016-05-19T05:02:18

Plug the coefficients into the quadratic formula and get the rather messy $x=-3/14+439/14i$ and $x=-3/14-439/14i$ as a solution

We'll be using the quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Let's substitute in:

$x=(3+-sqrt(-3^2-4(-7)(-16)))/(2(-7))$

and simplify:

$x=(3+-sqrt(9-448))/(-14)$
$x=(3+-sqrt(-439))/(-14)$

So the roots are:

$x=-3/14+sqrt(-439)/14$ and $x=-3/14-sqrt(-439)/14$

A negative under the square root makes that number imaginary, so we use the following symbology:

$x=-3/14+439/14i$ and $x=-3/14-439/14i$

How do you find the zeros, real and imaginary, of y= -7x^2-3x-16y= -7x^2-3x-16 using the quadratic formula?