If α and β are the zeroes of $3x^2-10x+2$, the value of α² + β² is:

$\alpha +\beta =-\frac{b}{a}=\frac{10}{3}$ and $\alpha \beta =\frac{c}{a}=\frac{2}{3}$. Then ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =\frac{100}{9}-\frac{4}{3}=\frac{100}{9}-\frac{12}{9}=\frac{88}{9}.\frac{100}{9}$ drops the $-2\alpha \beta$ term, $\frac{112}{9}$ adds that term instead of subtracting it, and $\frac{88}{3}$ forgets to square the denominator when squaring the sum.