Question

How do you find the zeros, real and imaginary, of `y= -x^2-5x-14` using the quadratic formula?

Answer 1

Two imaginary roots `(-(5+-sqrt(31)i)/2)`

Explanation:

For a quadratic equation in the general form
`color(white)("XXX")y=ax^2+bx+c`
the quadratic formula
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`
gives us the zeros (or "roots") of the equation.

For the specific equation
`color(white)("XXX")y=-x^2-5x-14`
`color(white)("XXXXXXXXXXXX")a=(-1)`
`color(white)("XXXXXXXXXXXX")b=(-5)`
`color(white)("XXXXXXXXXXXX")c=(-14)`
so the quadratic formula becomes
`color(white)("XXX")x=(-(-5)+-sqrt((-5)^2-4(-1)(-14)))/(2(-1))`

`color(white)("XXX")=(5+-sqrt(25-56))/(-2)`

`color(white)("XXX")=-(5+-sqrt(31)i)/2`