completing the square.......how do you solve?

Question

i hate doing these...well i never really finish...but i have to complete this square.....x^2-6x=-3....how do i do this?...thanks in advance!!

Answer 1

((x-3)^2)-9=-3
(x-3)^2=6
(x-3)=±√6
(x-3)=±2.4495
x=±2.4495+3
x=5.4495, 0.5505

Answer 2

The first part is to recognise what the square is going to be. For single x^2 - It is half of the 'x' value - in this case -6/2 = -3

So we have (x-3)^2 which equals x^2-6x + 9
You have x^2-6x + 3 = 0 which can be written as
(x^2-6x+ 9) - 6 = 0
So (x-3)^2 - 6 = 0 or (x-3)^2 = 6
Take square roots NB Plural

x-3 = +sqr(6) or x-3 = -sqr(6)

x = 3 + sqr(6)
x = 3 - sqr(6)

PS try completing the square of a general equation ax^2 + bx +c and see if you can get a general solution - hint divide by a to start.

Answer 3

x² - 6x + 9 = - 3 + 9
(x - 3)² = 6
(x - 3) = ±√6
x = 3 ± √6

Answer 4

x^2 - 6x = - 3
x^2 - 6x + 9 = 9 - 3
(x - 3)^2 = 6
x - 3 = ±√6
x = 3 -√6, 3 +√6