Solve by completing the square of 3x2-8x-6=0?

Question

The problem is 3x(squared) - 8x - 6 = 0

Choices for the answer:

(-4+/-(sqaure root)34) / 3 (meaning the first part is over 3)

(4+/-(square root)34) / 3

-4+/-2(square root)34

4+/-2(square root)34

Answer 1

1. Move the constant to the right side of the equation:
3x^2 - 8x = 6

2. Divide out the x^2 coefficient:
x^2 - 8/3x = 2

3. Divide the coefficient of x (in this case, -8/3) by 2 and square it, then add it to both sides:
x^2 - 8/3x + 16/9 = 34/9

4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:
(x - 4/3)^2 = 34/9

5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:
x - 4/3 = +/-sqrt(34/9)

6. Move the constant on the left to the right and simplify:
x = 4/3 +/-sqrt(34/9) = 4/3 +/-(sqrt 34)/3 = (4 +/- sqrt 34)/3

Answer 2

Let's put the 6 on the other side.

3x^2 - 8x = 6

Now make the left side a perfect square.

3x^2 - 8x + 4/3 = 6 + 4/3

Answer 3

alcohol