Question

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

Answer 1

Expand the expression on the right side into standard form, then apply the quadratic formula.
In this case there will be two imaginary roots `2+-isqrt(5)`

Explanation:

`y=(x-3)^2+2`

`rArr y= 1x^2+(-4)x+9`

which is an example of quadratic standard form: `y=ax^2+bx+c`
for which the roots are given by the quadratic formula
`color(white)("XXX")x= (-b+1sqrt(b^2-4ac))/(-2a)`

Which for our example becomes:
`color(white)("XXX")x= (-(-4)+-sqrt((-4)^2-4(1)(9)))/(2(1))`

`color(white)("XXXX")= (4+-sqrt(16-36))/2`

`color(white)("XXXX")=(4+-2sqrt(-5))/2`

`color(white)("XXXX")=2+-isqrt(5)`