solve for x

3x^2-12x+3=0

is the answer 1/3, 3

according to your solutions, the equation would factor to

(3x-1)(x-3) = 0

would this give you the original equation?
Clearly not!

so , like I asked before, how are you getting your answers?
What method are you using?

I tried it your way and this is what I came up with

x = 2�} 6�ã3

am I right?

use the formula

x = (-b ±√(b^2-4ac)/(2a)
= (12 ± √108)/6
= (12 ±6√3)/6
= 2 ± √3

To solve the quadratic equation 3x^2 - 12x + 3 = 0, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this equation, a is the coefficient of x^2 (which is 3), b is the coefficient of x (which is -12), and c is the constant term (which is 3).

Substituting these values into the quadratic formula, we get:

x = (-(-12) ± sqrt((-12)^2 - 4(3)(3))) / 2(3)
x = (12 ± sqrt(144 - 36)) / 6
x = (12 ± sqrt(108)) / 6
x = (12 ± sqrt(36 * 3)) / 6
x = (12 ± 6sqrt(3)) / 6
x = (2 ± sqrt(3))

So the solutions to the equation 3x^2 - 12x + 3 = 0 are x = 2 + sqrt(3) and x = 2 - sqrt(3).
Therefore, the answer is x = 2 + sqrt(3) and x = 2 - sqrt(3).