Question

How do you solve the equation `3x^2-24x+27=0` by completing the square?

Answer 1

You may start by dividing everything by `3`

Explanation:

`->x^2-8x+9=0`

You take half of the `x` coefficient and square that. The square would then be:
`x^2-8x+(-4)^2=x^2-8x+16=(x-4)^2`

To get there you would have added `7` to the original `9`. Also add `7` to the other side of the `=` sign:

`->(x-4)^2=7->x-4=+-sqrt7->x=4+-sqrt7`