Question

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

Answer 1

`x = (5 +- i*sqrt(2))/3`

Explanation:

First, square `(x-2)` and combine like terms.
`y = 3(x-2)^2 + 2x - 3`
`y = 3(x-2)(x-2) + 2x - 3`
`y = 3(x^2 - 4x + 4) + 2x - 3`
`y = 3x^2 - 12x + 12 + 2x - 3`
`y = 3x^2 - 10x + 9`

Quadratic Formula:
`x = (-b +- sqrt(b^2 - 4ac))/(2a)`
`x = (-(-10) +- sqrt((-10)^2 - 4(3)(9)))/(2(3))`
`x = (10 +- sqrt(100 - 108))/6`
`x = (10 +- sqrt(-8))/6`
`x = (10 +- 2i*sqrt(2))/6`

Simplify:
`x = (5 +- i*sqrt(2))/3`