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.
 A: $$3x^2-2x+7=0\Longleftrightarrow$$
$$x=\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot 3 \cdot 7}}{2\cdot 3}\Longleftrightarrow$$
$$x=\frac{2\pm\sqrt{4-4\cdot 3 \cdot 7}}{6}\Longleftrightarrow$$
$$x=\frac{2\pm\sqrt{4-84}}{6}\Longleftrightarrow$$
$$x=\frac{2\pm\sqrt{-80}}{6}\Longleftrightarrow$$
$$x=\frac{2\pm i\sqrt{80}}{6}\Longleftrightarrow$$
$$x=\frac{2\pm 4i\sqrt{5}}{6}\Longleftrightarrow$$
$$x=\frac{2 + 4i\sqrt{5}}{6} \vee x=\frac{2 - 4i\sqrt{5}}{6}\Longleftrightarrow$$
$$x=\frac{2 + 4i\sqrt{5}}{6} \vee x=\frac{2 - 4i\sqrt{5}}{6}$$

$$\left(\left(\frac{2 + 4i\sqrt{5}}{6}\right)-\frac{1}{3}\right)^2 =\left(\frac{2i\sqrt{5}}{3}\right)^2 =\frac{4i^2\cdot 5}{9}=-\frac{20}{9}$$
$$\left(\left(\frac{2 - 4i\sqrt{5}}{6}\right)-\frac{1}{3}\right)^2 =\left(\frac{-2i\sqrt{5}}{3}\right)^2=\frac{4i^2\cdot 5}{9}=-\frac{20}{9}$$
So as we see the answer is $\color{red}{-\frac{20}{9}}$
A: HINT:
complete the square in $$3x^2-2x+7$$
A: Starting from
$$3x^2-2x+7=0\\$$
$$x^2-\frac{2}{3}x+\frac{7}{3}=0\\$$
$$x^2-2\cdot\frac{1}{3}\cdot x+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2+\frac{7}{3}=0\\$$
$$\left(x-\frac{1}{3}\right)^2+\frac{21-1}{9}=0\\$$
$$ \left(x-\frac{1}{3}\right)^2=\frac{-20}{9}$$
A: Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+\frac{1}{3}+7-\frac{1}{3}=0$$ $$3\left(x^2-\frac{2x}{3}+\frac{1}{9}\right)+7-\frac{1}{3}=0$$ $$3\left(x-\frac{1}{3}\right)^2+\frac{21-1}{3}=0$$
$$3\left(x-\frac{1}{3}\right)^2=-\frac{20}{3}$$
$$\left(x-\frac{1}{3}\right)^2=-\frac{20}{9}$$
Hence, we get 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\left(x-\frac{1}{3}\right)^2=\color{blue}{-\frac{20}{9}}}}$$
A: 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
$$
\left(x-\frac{1}{3}\right)^2=\frac{1}{3}\left(3x^2-2x+7-7\right)+\frac{1}{9}$$ $$=-\frac{7}{3}+\frac{1}{9}
$$ $$=-\frac{20}{9}
$$
A: HINT:
$$3x^2-2x+7=3\left(x-\frac13\right)^2+6+\frac23$$
A: Starting from, $$3x^2-2x+7=0$$ Dividing by 3:$$x^2-\frac{2x}{3}+\frac{7}{3}=0$$ Completing the square: $$\left(x-\frac 13\right)^2-\frac{1}{9}+\frac{7}{3}=0$$  $$\left(x-\frac 13\right)^2=\frac{1}{9}-\frac{7}{3}$$ $$\left(x-\frac 13\right)^2=\frac{1-21}{9}$$ 
$$\left(x-\frac 13\right)^2=-\frac{20}{9}$$ 
A: Expand the binomial and compare it to the trinomial.
$$3x^2-2x+7\iff\left(x-\frac13\right)^2=x^2-\frac23x+\frac19.$$
If you divide the polynomial by $3$, you get closer, with two identical terms
$$\frac{3x^2-2x+7}3=x^2-\frac23x+\frac73.$$
To get a perfect identity, it now suffices to add a well-chosen constant
$$\frac{3x^2-2x+7}3-\frac{20}9=x^2-\frac23x+\frac19.$$
Now as the polynomial is known to evaluate to $0$, you know the value of the RHS.
