We have the quadratic equation 3x^2 -5x -4 = 0.

The roots of the equation are

x1 = [-b + sqrt ( b^2 - 4ac)]/2a

=> [ 5 + sqrt ( 25 + 48)]/ 6

=> 5/6 + sqrt 73 / 6

x2 = [-b - sqrt ( b^2 - 4ac)]/2a

=> [ 5 - sqrt ( 25 + 48)]/ 6

=> 5/6 - sqrt 73 / 6

**Therefore the roots are 5/6 + sqrt 73 / 6 and 5/6 - sqrt 73 / 6.**

Given the quadratic equation:

3x^2 - 5x - 4 = 0

==> a = 3 b= -5 c = -4

We need to solve for x using the quadratic formula.

We know that:

x = (-b +- sqrt(b^2 -4ac) / 2a

==> x1= ( 5 + sqrt(25+4*3*4) /2*3

= (5 + sqrt(73) / 6

**==> x1= (5+sqrt73)/6**

**==> x2= (5-sqrt73)/6**

The roots x1 and x2 of the quadratic equation ax^2+bx+c = 0 is given by the formula:

x1 = {-b+sqrt(b^2-4ac)}/2 and

x2 = -{b+sqrt(b^2-4ac)}/2.

The given quadratic equation is 3x^2-5x-4 = 0.

So a = 3, b= -5 and c= -4.

So using the formula we get the two roots:

x1 = {-(-5)+sqrt(-5)^2-4*3(-4)}/2*3 = {5+sqrt(25+48)}/6

x1 = {5+sqrt(73)}/6.

x2 = {5-sqrt(73)}/6.