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

Mar 20, 2018

There are no zeros for the function specified.

#### Explanation:

I first attempted to solve this using the quadratic formula:

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

However, the $4 a c$ 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'.