Is this how you solve t(t+1)(t+2)=0, x^3+4x=4x^2, and 3r^2=4r-1?

Question

t(t+1)(t+2)=0
t=0 or t= -1 or t=2
{0, -1, -2}

x^3+4x=4x^2
x^3+4x-4x^2=0
x(x^2+4-4x)=0
x(x^2-4x+4)=0
x(x-2)(x-2)=0
x=0 or x=2
{0,2(double root)}

3r^2=4r-1
3r^2-4r+1=0
r^2-4r+3=0
(r-3)(r-1)=0
r=3 or r=1
{1,3}

Answer 1

The first two are fine.

The third one: 3r^2 - 4 + 1 = (3r-1)(r-1)

Take it from there.

Answer 2

thanks

it was {1/3, 1} right