I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$ \begin{align} 3x^2 - 4x -2 = 0 \\ 3x^2 - 4x = 2 \end{align} $$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$ \begin{align} x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\ \left(x - \frac{2}{3}\right)^2 = \frac{10}{9} \end{align} $$
Can anyone please explain how they went from the first step to the second step?




Try going backward; expand the square $(x-\tfrac{2}{3})^2$ to find that $$\left(x-\frac{2}{3}\right)^2=\left(x-\frac{2}{3}\right)\left(x-\frac{2}{3}\right)=x^2-\frac43x+\frac49.$$