Solving this quadratic equation by completing the square? $-\frac{2}{3}x^{2} - x +2 = 0$
Here's what I did:

However, the textbook answer is $x = -2.6, x = 1.1$. What did I do wrong?
 A: $$-\frac{2}{3}\bigg(x^2+\frac{3}{2}x+\frac{9}{16} \bigg)+\frac{19}{8}$$
is the equivalent of this expression in a square form. Here notice that constant term should be equal to $2$, therefore we add $\dfrac{19}{8}$ in order to have $-\dfrac{2}{3} \cdot \dfrac{9}{16} + \dfrac{19}{8} = 2$. 
For the rest of the question, your logic is completely correct so I leave it to you.
A: \begin{align*}
-\frac{2}{3}x^2 - x + 2 & = 0\\
-\frac{2}{3}\left(x^2 + \frac{3}{2}x\right) + 2 & = 0\\
-\frac{2}{3}\left[x^2 + \frac{3}{2}x + \left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right] \color{red}{- \left[-\frac{2}{3}\left(\frac{1}{2} \cdot \frac{3}{2}\right)^2\right]} + 2 & = 0 \tag{1}\\
-\frac{2}{3}\left(x^2 + \frac{3}{2}x + \frac{9}{16}\right) + \frac{3}{8} + 2 & = 0\\
-\frac{2}{3}\left(x + \color{red}{\frac{3}{4}}\right)^2 + \frac{19}{8} & = 0 \tag{2}\\
-\frac{2}{3}\left(x + \frac{3}{4}\right)^2 & = -\frac{19}{8}\\
\left(x + \frac{3}{4}\right)^2 & = \frac{57}{16}\\
\left|x + \frac{3}{4}\right| & = \sqrt{\frac{57}{16}}\\
x + \frac{3}{4} & = \pm \frac{\sqrt{57}}{4} \tag{3}\\
x & = -\frac{3}{4} \pm \frac{\sqrt{57}}{4}\\
x & = \frac{-3 \pm \sqrt{57}}{4}
\end{align*}
(1):  If you add a term to one side of the equation, you must either add it to the other side of the equation or subtract it from the same side of the equation to balance the equation.  You omitted the term in red.
(2):  Note that 
$$\frac{9}{16} = \left(\frac{3}{4}\right)^2$$
You made the mistake of treating $9/16$ as the square of $3/2$.
(3):  Since $16$ is a perfect square, you do not need to rationalize the denominator in this step.  What you did was correct, but introducing extra steps introduces extra opportunities to make an error.
A: This is not an answer, but too long for a comment.
Make your life easy by writing
$$-\frac23x^2-x+2=0\iff x^2+\frac32x-3=0.$$
Then
$$\left(x+\frac34\right)^2-\left(\frac34\right)^2-3=0,$$
$$\left(x+\frac34\right)^2=\frac{57}{16},$$
$$x+\frac34=\pm\frac{\sqrt{57}}4.$$
A: You're wrong in this step:
$-\cfrac{2}{3}\biggl(x^2+\cfrac{3}{2}x+\cfrac{9}{16}\biggr)+2+\cfrac{3}{8}=0$
$-\cfrac{2}{3}\biggl(x^2+\cfrac{3}{2}\biggr)^2+\cfrac{19}{8}=0$  $\leftarrow$ Here
The actual result of that is:
$-\cfrac{2}{3}\biggl(x^2+\cfrac{3}{4}\biggr)^2+\cfrac{19}{8}=0$
