Let the mean and the variance of seven observations 2, 4, α, 8, β,...

Given the mean of the seven observations is $8$:

$\dfrac{2 + 4 + \alpha + 8 + \beta + 12 + 14}{7} = 8$

$\dfrac{40 + \alpha + \beta}{7} = 8$

$\alpha + \beta = 16$

Given the variance of the observations is $16$:

$\dfrac{\sum x_i^2}{n} - (\text{Mean})^2 = 16$

$\dfrac{2^2 + 4^2 + \alpha^2 + 8^2 + \beta^2 + 12^2 + 14^2}{7} - 8^2 = 16$

$\dfrac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 64 = 16$

$\dfrac{424 + \alpha^2 + \beta^2}{7} = 80$

$424 + \alpha^2 + \beta^2 = 560$

$\alpha^2 + \beta^2 = 136$

Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$:

$16^2 = 136 + 2\alpha\beta$

$256 = 136 + 2\alpha\beta$

$2\alpha\beta = 120 \Rightarrow \alpha\beta = 60$

Solving for $\alpha$ and $\beta$, they are the roots of the equation $t^2 - 16t + 60 = 0$, which are $6$ and $10$.

Since $\alpha < \beta$, we have $\alpha = 6$ and $\beta = 10$.

The roots of the required quadratic equation are $3\alpha + 2$ and $2\beta + 1$:

$3\alpha + 2 = 3(6) + 2 = 20$

$2\beta + 1 = 2(10) + 1 = 21$

Sum of the roots $= 20 + 21 = 41$

Product of the roots $= 20 \times 21 = 420$

The required quadratic equation is $x^2 - (\text{Sum})x + (\text{Product}) = 0$:

$x^2 - 41x + 420 = 0$

Answer: $x^2 - 41x + 420 = 0$