Answer
$x=2$ 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:
$3(x^{2}-\dfrac{4}{3}x)=4$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to the expression inside the parentheses and $3\Big(\dfrac{b}{2}\Big)^{2}$ to the right side of the equation. In this particular problem, $b=-\dfrac{4}{3}$
$3\Big[x^{2}-\dfrac{4}{3}x+\Big(-\dfrac{2}{3}\Big)^{2}\Big]=4+3\Big(-\dfrac{2}{3}\Big)^{2}$
$3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=4+3\Big(\dfrac{4}{9}\Big)$
$3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=4+\dfrac{4}{3}$
$3\Big(x^{2}-\dfrac{4}{3}x+\dfrac{4}{9}\Big)=\dfrac{16}{3}$
Factor the expression inside the parentheses, which is a perfect square trinomial:
$3\Big(x-\dfrac{2}{3}\Big)^{2}=\dfrac{16}{3}$
Take the $3$ to divide the right side of the equation:
$\Big(x-\dfrac{2}{3}\Big)^{2}=\dfrac{16}{9}$
Take the square root of both sides of the equation:
$\sqrt{\Big(x-\dfrac{2}{3}\Big)^{2}}=\sqrt{\dfrac{16}{9}}$
$x-\dfrac{2}{3}=\pm\dfrac{4}{3}$
Solve for $x$:
$x=\dfrac{2\pm4}{3}$
The two solutions are:
$x=\dfrac{2+4}{3}=2$
$x=\dfrac{2-4}{3}=-\dfrac{2}{3}$
