If α and β are the zeroes of $3x^2-11x+6$, the value of $(\alpha -\beta {)}^2$ is:

$\alpha +\beta =\frac{11}{3}$ and $\alpha \beta =\frac{6}{3}=2.(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta =\frac{121}{9}-8=\frac{121}{9}-\frac{72}{9}=\frac{49}{9}$. (Zeroes 3 and $\frac{2}{3}$ differ by $\frac{7}{3}$, and ${\left(\frac{7}{3}\right)}^2=\frac{49}{9}.)\frac{121}{9}$ drops the $-4\alpha \beta$ term; $\frac{85}{9}$ subtracts $2\alpha \beta$, which is the ${\alpha }^2+{\beta }^2$ formula, not this one; $\frac{7}{3}$ is the difference itself, not its square.