help 3x^2-2=6x?

Question

solve by completing the square
can anyone help me solve this?
thanks

Answer 1

Get the x² and x terms on one side, number on the other:

3x² - 6x = 2

Now divide by 3 so you have x² with no coefficient:

x² - 2x = 2/3

Now you can complete the square. Take the coefficient on the x term (-2), halve it (-1) and square it (1). Now add that to both sides:

x² - 2x + 1 = 1 + 2/3
x² - 2x + 1 = 5/3

The expression on the left is now a perfect square so rewrite it as such:
(x - 1)² = 5/3

Take the square root of both sides (don't forget ±)
x - 1 = ± sqrt(5/3)

Add 1 to both sides:
x = 1 ± sqrt(5/3)

Answer 2

3x^2-2=6x
3x^2-6x= 2

Divide throughout by 3
==> x^2 - 2x = 2/3
Take half the coefficient of x, square it and add it to both side

==> x^2 - 2x + (-1)^2 = 2/3+ (-1)^2
Factorise LHS

==> (x-1)^2 =2/3 + 1

Take square root on both sides
x-1 = +or - (5/3)^1/2

x = 1+ (5/3)^1/2 or
x = 1- (5/3)^1/2

Answer 3

3x^2-2=6x

3x^2-6x=2
Completing the square:
3(x^2-2x+1) = 2 + 3 3 is from the left (3x1)
Simplify:
3(x-1)^2 = 5

(x-1)^2 =5/3

x-1=square root of 5/3

x = 1 + square root of 5/3 or 2.29

Answer 4

3x^2 -- 2 = 6x
=> 3x^2 -- 6x -- 2 = 0
=> 3(x^2 -- 2x + 1) = 5
=> (x -- 1)^2 = 5/3
=> x -- 1 = sqrt(5/3)
=> x = 1 + sqrt(5/3)

Answer 5

3x² - 6x - 2 = 0
3(x² - 2x - 2/3) = 0
x² - 2x - 2/3 = 0
(x² - 2x + 1) - 1 - 2/3 = 0
(x - 1)² = 5/3
(x - 1) = ±√ (5/3)
x = 1 ± √(5/3)