Let the mean and the variance of seven observations \(2,4, \alpha, 8, \beta,12,14, \alpha<\beta\), be \(8\) and \(16\…

Mean \(\bar{x}=\frac{\sum x_i}{n}\)

\(8=\frac{2+4+\alpha+8+\beta+12+14}{7}\)

\(56=40+\alpha+\beta\)

\(\alpha+\beta=16\)

and \(\sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2\)

\(16=\frac{4+16+\alpha^2+64+\beta^2+144+196}{7}-64\)

\(560=424+\alpha^2+\beta^2\)

\(\alpha^2+\beta^2=136\)

Since \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\)

\(256=136+2 \alpha \beta\)

\(\alpha \beta=60\)

\(\alpha\) and \(\beta\) are the roots of the quadratic equation \(t^2-(\alpha+\beta) t+\alpha \beta=0\)

\(t^2-16 t+60=0\)

\((t-6)(t-10)=0\)

\(\alpha=6\) and \(\beta=10\)

Now, \(3 \alpha+2=20\) and \(2 \beta+1=21\)

Sum of roots \(=20+21=41\)

Product of roots \(=420\)

Required equation is

\(x^2-41 x+420=0\)