Question

How do you find the roots, real and imaginary, of `y=(x +5 )^2+6x+3` using the quadratic formula?

Answer 1

`x1 = -14`
`x2 = -2`

Explanation:

`y = (x + 5)^2 +6x +3`
Expand square term: `y = x^2 + 10x + 25 + 6x + 3`
Therefore `y = x^2 +16x + 28`

This quadratic factorizes so there is no need of the Quadratic Formula:

`y = (x + 14)(x + 2)` which has zeros at `x = -14 and -2`

If you would like to use the Quadratic Formula:

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

In this case `a = 1, b = 16, c = 28`

Hence; `x = (-16 +- sqrt(256 - 112))/2`
`x = (-16+- sqrt(144))/2`
`x= (-16 +- 12)/2`

`x = -14 or -2`

Regarding your question on real or complex roots, the roots of a quadratic function with real coefficients will either both be real or both be complex. In this case the roots are real.