Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice: 
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$. 
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sure where to go from here. What I tried was the following:
$(1/3)|x+3| < 2$, so $|x+3|<6$
Since the maximum of $x$ is $7/3$, the most $|x+3|$ can be is $|(7/3)+3|= 16/3 < 6$.
Then, the minimum of $x$ is $5/3$, so the least $|x+3|$ can be is $|(5/3)+3| = 14/3 < 6$.
Thus, $|x-2||x+3| < 2$.
Anyone know if this is correct? If not, can you point me in the right direction? Thanks in advance.

Let $x \in \mathbb R$ and suppose that $|x−2| < \dfrac 13$. Prove that $|x^2+x−6| < 2$.
\begin{align}
   |x-2| < \dfrac 13 
    &\implies \dfrac 53 < x < \dfrac 73 \\
    &\implies 3+\dfrac 53 < x+3 < 3 + \dfrac 73 \\
    &\implies -\dfrac{16}{3} < x+3 < \dfrac{16}{3} \\
    &\implies |x+3| < \dfrac{16}{3} \\
\end{align}
Hence $|x^2 + x - 6| = |x-2||x+3| < \dfrac 13 \cdot \dfrac{16}{3} < \dfrac{16}{9} < 2$.
