Answer
The solutions are $x=2$ and $x=-\dfrac{2}{3}$
Work Step by Step
$3x^{2}-4x=4$
Take out common factor $3$ from the left side of the equation:
$3\Big(x^{2}-\dfrac{4}{3}x\Big)=4$
Take $3$ to divide the right side:
$x^{2}-\dfrac{4}{3}x=\dfrac{4}{3}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=-\dfrac{4}{3}$:
$x^{2}-\dfrac{4}{3}x+\Big(-\dfrac{4}{3\cdot2}\Big)^{2}=\dfrac{4}{3}+\Big(-\dfrac{4}{3\cdot2}\Big)^{2}$
$x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}=\dfrac{4}{3}+\dfrac{4}{9}$
$x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}=\dfrac{16}{9}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{2}{3}\Big)^{2}=\dfrac{16}{9}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{2}{3}\Big)^{2}}=\pm\sqrt{\dfrac{16}{9}}$
$x-\dfrac{2}{3}=\pm\dfrac{4}{3}$
Solve for $x$:
$x=\pm\dfrac{4}{3}+\dfrac{2}{3}$
$x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2$
$x=-\dfrac{4}{3}+\dfrac{2}{3}=-\dfrac{2}{3}$
The solutions are $x=2$ and $x=-\dfrac{2}{3}$
