Solving $3t^2-\frac{12}{3}t+\frac{4}{3}=0$ I need to to solve:
$$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$
The solution manual factorizes this to $\dfrac{1}{3}(3t-2)^2$. How can you do this easily?

A problem like this is easy if you make a substitution, for example.
$$3t^2-\frac{12}{3}t+\frac{4}{3}=0$$
$$\frac{9}{3}t^2-\frac{12}{3}t+\frac{4}{3}=0$$
$$\frac{1}{3}(9t^2-12t+4)=0$$
$$\frac{1}{3}((3t)^2-4(3t)+4)=0$$
Now, let $s=3t$
$$\frac{1}{3}(s^2-4s+4)=0$$
$$\frac{1}{3}(s-2)^2=0$$
Substituting back we have
$$\frac{1}{3}(3t-2)^2=0$$
