I tried doing it but I got 4. How do I begin to solve this problem?
We are given:
$\displaystyle 3x^2-2x+7=0$
The first thing I would do is subtract through by 7, then divide through by 3 to get:
$\displaystyle x^2-\frac{2}{3}x=-\frac{7}{3}$
Next, complete the square on the left side by adding the square of one half the coefficient of the linear term, i.e. add $\displaystyle \left(\frac{1}{2}\cdot\frac{2}{3} \right)^2=\frac{1}{9}$ to get:
$\displaystyle x^2-\frac{2}{3}x+\frac{1}{9}=-\frac{7}{3}+\frac{1}{9}$
Rewrite the left side as the square of a binomial and combine like terms on the right, and what do you get?