Highschool Inequality For all $a>0$ find a $b>0$ such that 
$|t-2|<b\Longrightarrow\sqrt{(t^2-4)^2+(3t-6)^2}<a$
$\textbf{Proof:}$
Let us square both sides of the second inequality, then we have
$(t^2-4)^2+(3t-6)^2<a^2\Longrightarrow (t-2)^2(t^2+2t+13)<a^2$
That is as far I can go

We have
$$
\sqrt{(t^2-4)^2+(3t-6)^2}=\sqrt{|t-2|^2 ((t+2)^2+3^2)}< b \sqrt{(t+2)^2+3^2}.
$$
Moreover by the triangular inequality the last term is $\le$ than
$$
b \sqrt{(|t-2|+4)^2+3^2}<b \sqrt{\left(b^2+8b+9\right)+16}.
$$
Therefore it would be sufficient that
$$
b\sqrt{b^2+8b+25}<a,
$$
which is equivalent to
$$
b^2(b^2+8b+16)<a^2.
$$
Obviosly the LHS is increasing in $b$ [with $b>0$], therefore it is sufficient to choose a sufficiently small $b>0$.
