Can't find my mistake

Question

I'm trying to find the sum of the reciprocal numbers of squares of quadratic equation:$3x^2-14x+6=0$, I managed to find the answer by calculating the roots, and summing their reciprocal. However firstly I tried to do it using viet's formulas here is what I tried:
$\frac{1}{x_1}+ \frac{1}{x_2}= $
$\frac{x_2+x_1}{2x_1x_2}= $ (Commond dominator)
$\frac{1}{2}*\frac{x_1+x_2}{x_1*x_2}=$
$\frac{1}{2}*\frac{\frac{-b}{a}}{\frac{c}{a}}=$
$\frac{1}{2}*\frac{-b}{c}=$
$\frac{1}{2}*\frac{14}{6}=$
$\frac{14}{12}=\frac{7}{6}$ However the correct answer is $\frac{7}{3}$. What did I missed? I'm assuming that $a+b=b+c$ and $a*b=b*a$

Answer 1

You have incorrectly added the fractions - you wouldn't multiply the denominator by 2 when using the common denominator! You should simply have -b/c = 14/6 = 7/3. Voila!!! Hope this helps!

Answer 2

$$\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1x_2}\ne\frac{x_1+x_2}{\color{red}{2}x_1x_2}$$