The roots of quadratic equation $3x^2-2x+\frac{1}{3}=0$ are  

Given quadratic equation 

$3x^2-2x+\frac{1}{3}=0$.............(i) 

On comparing eq (i) with $a{x}^2+bx+c=0$ we get

$a=3, b=-2, c=\frac{1}{3}$ 

Discriminant of quadratic equation 

$D=b^2-4ac$ 

Putting the value of a, b, c 

$D={(-2)}^2-4\times 3\times \frac{1}{3}$ 

$D=4-4$

$D=0$

Hence the roots of given quadratic equation are real and equal.