Question

What are the zero(s) `1x^2-6x+20=0`?

Answer 1

There are no zeros for the function specified.

Explanation:

I first attempted to solve this using the quadratic formula:

`(-b+-sqrt(b^2-4ac))/(2a)`

However, the `4ac` term ends up being much larger than `b^2`, making the term under the radical negative and therefore imaginary.

My next thought was to plot and just check if the graph crosses the x-axis:

graph{x^2-6x+20 [-37.67, 42.33, -6.08, 33.92]}

As you can see, the plot doesn't cross the x-axis, and therefore has no 'zeros'.