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.

Starting from
$$3x^2-2x+7=0\\$$
$$x^2-\frac{2}{3}x+\frac{7}{3}=0\\$$
$$x^2-2\cdot\frac{1}{3}\cdot x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{7}{3}=0\\$$
$$\left(x-\frac{1}{3}\right)^2+\frac{21-1}{9}=0\\$$
$$ \left(x-\frac{1}{3}\right)^2=\frac{-20}{9}$$
