Question

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

Answer 1

In general `ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))`

For your problem, `a=3`, `b=-2` and `c=-12`, so we get

`0 = ax^2+bx+c = 3x^2 - 2x - 12`

`= 3(x+(-2)/(2*3))^2+(-12-(-2)^2/(4*3))`

`= 3(x-2/6)^2-(12+4/12)`

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

`= 3(x-1/3)^2-37/3`

Adding `37/3` to both sides we get

`3(x-1/3)^2 = 37/3`

Dividing both sides by 3 we get

`(x-1/3)^2 = 37/(3^2)`

Hence

`x-1/3 = +- sqrt(37/3^2) = +- sqrt(37)/3`

Adding 1/3 to both sides, we get

`x = 1/3 +- sqrt(37)/3 = (1 +- sqrt(37))/3`

Answer 2

If `3x^2-2x-12 = 0`
then
`x^2-2/3x -4 = 0`

isolate the constant
`x^2-2/3x = 4`

complete the square
`x^2 - 2/3x + (1/3)^2 = 4 + (1/3)^2`

`(x-1/3)^2 = 37/9`

take the square root
`x-1/3 = +-sqrt(37)/3`

simplify
`x= (1+-sqrt(37))/3`