Question

How do you find the zeros, real and imaginary, of `y= x^2-5x+16` using the quadratic formula?

Answer 1

`x=(5+isqrt(39))/(2)` and `x=(5-isqrt(39))/(2)`

Explanation:

The quadratic formula is: `(-b+-sqrt(b^2-4ac))/(2a)`.

This is used to solve equations of the form `y=ax^2+bx+c`, which is the same type of equation as `y=x^2-5x+16`. We can see that for our example, `a=1`, `b=-5`, `c=16`. Plugging these guys into the formula,
`(-(-5)+-sqrt((-5)^2-4(1)(16)))/(2(1))`

And simplifying,
`(5+-sqrt(25-64))/(2)`

`(5+-sqrt(-39))/(2)`

We now see we have an issue - a negative in the square root. That means both of our solutions will be imaginary, so to represent that, we take the `sqrt(-1)` (`i`) out of the square root to make it positive:
`(5+-isqrt(39))/(2)`

The next thing we look for is any way to simplify the square root. But `sqrt(39)` can't be simplified further, which means our solutions are:
`x=(5+isqrt(39))/(2)`
`x=(5-isqrt(39))/(2)`