P(x)=3x^3-10x^2-5x has one root -2/3. What are the other roots?

P(x)=3x^3-10x^2-5x.

The three
roots of this equation can be found by equating  3x^3-10x^2-5x to
0.

=> 3x^3-10x^2-5x
=0

=> x ( 3x^2 - 10x - 5) =
0

So one of the roots is 0

The
other two roots are

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

=> x2 = [10 + sqrt (100 +
60)]/6

=> x2 = 10/6 + (sqrt 160) /
6

x3 = [-b - sqrt (b^2 -
4ac)]/2a

=> x3 = [10 - sqrt (100 +
60)]/6

=> x3 = 10/6 - (sqrt 160) /
6

Therefore the roots are 0, 10/6 + (sqrt
160) / 6 and 10/6 - (sqrt 160) / 6.