<s>[INST] If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$  then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand it now.
 [/INST] Observe $$\left(x-\frac{1}{3}\right)^2=x^2-\frac{2}{3}x+\frac{1}{9}$$$$=\frac{1}{3}\left(3x^2-2x\right)+\frac{1}{9}.$$
This is almost the original expression, we're just missing a $7$.  Then,
$$\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}.$$
Now, use the original equality to simplify.
Then, we get
$$
\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}$$ $$=-\frac{7}{3}+\frac{1}{9}
$$ $$=-\frac{20}{9}
$$
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