Question

How do you solve by completing the square for `3x^2 = 6x + 4`?

Answer 1

`f(x) = 3x^2 - 6x - 4` = 0

f(x) = 3(x^2 - 2x - 4/3) = 3(x^2 - 2x + 1) - 3 - 12/3

(x - 1)^2 = 21/9 =

`(x - 1) = +- (sqrt21)/3` ->

`(x - 1) = (sqrt21)/3 -> x = 1 + (sqrt21)/3`

`(x - 1) = -(sqrt21)/3 -> x = 1 - (sqrt21)/3`