Question

How do you solve by completing the square `3x^2 + 6x +12 = 0`?

Answer 1

First divide both sides of the equation by the coefficient of `x^2` (in this case `3`)
`x^2+2x+4=0`

If `x^2 + 2x` are the first 2 terms of an expression from
`(x+a)^2`
then `a=1`

Rewrite the left-side of the expression as
`(x^2+2x+1) + (3) = 0`

`(x+1)^2 = -3`

This can not be solved with any real number value for `x`
as can be seen from the graph of the original equation below (it never is equal to `0`)
graph{3x^2+6x+4 [-16.03, 16, -1.31, 14.71]}

However, if we allow complex solutions
`x+1 = +-sqrt(-3)`

`x = -1 -sqrt(3)i` and `x=-1+sqrt(3)i`