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#