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

Finding the zeroes (or roots) of a quadratic equation means solving for `x` when `y=0`. In other words, we want to know what values of `x` we would have to plug into `x^2-x+7` in order for it to simplify to `0.`

To do this, there are some shortcuts that might be easy to see, but the quadratic formula is always an option. This formula states that, if we want `ax^2+bx+c` to equal `0,` then the `x`-values that make this true will be

`x=(–b+-sqrt(b^2-4ac))/(2a)`

So, for any specific quadratic expression (like `x^2-x+7`), all we need to do is plug the given values for `a, b,` and `c` into the quadratic formula.

For `x^2-x+7`, we have

`a=1` (from the `1x^2`),
`b=–1` (from the `-1x`), and
`c=7` (from the `+7`).

Plugging these into the quadratic formula gives

`x=(–color(purple)b+-sqrt(color(blue)b^2-4color(orange)acolor(green)c))/(2color(orange)a)`

`x=(–color(blue)((–1))+-sqrt(color(blue)((–1))^2-4color(orange)((1))color(green)((7))))/(2color(orange)((1)))`

`x=(1+-sqrt(1-28))/(2)`

`x=(1+-sqrt(–27))/(2)`

`x=(1+-3sqrt(–3))/(2)`

Because we get a negative under the square root, the zeroes are imaginary. Recalling that `i^2=–1`, we get

`x=(1+-3isqrt(3))/(2)`.