Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.

Because if the roots of $ax^2+bx+c$ are $r_1$ and $r_2$, then$$ax^2+bx+c=a(x-r_1)(x-r_2).$$You forgot the $a$.
In your case, you get:\begin{align}3x^2+4x-4&=3\left(x-\frac23\right)(x+2)\\&=(3x-2)(x+2).\end{align}
