<s>[INST] Solving $3x^2 - 4x -2 = 0$ by completing the square 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?
 [/INST] We get from $$3x^2-4x-2=0$$ to $$3x^2-4x=2$$ by adding two on both sides. Then dividing by $3$ we obtain
$$x^2-\frac{4}{3}x=\frac{2}{3}$$ then we can write $$x^2-2\cdot \frac{2}{3}x+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$
</s>