Is it possible to take the absolute value of both sides of an equation? I have a problem that says: Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$. 
Since the quadratic is greater than 0, I assume that there are no real solutions since 
$y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$ 
since $y>0$ there are no x-intercepts. I would use the discriminant $b^2-4ac<0$.
I now have $b^2-4(3)(7)<0$
$b^2-84<0$
$b^2<84$
$b<\pm\sqrt{84}$
Now how do I change $b$ to $|b|$? Can I take the absolute value of both sides of the equation or is there a proper way to do this? 

From $b^2\lt 84$, you cannot conclude that $b\lt \pm\sqrt{84}$, whatever that may mean. It cannot mean that $b\lt \sqrt{84}$ or $b\lt -\sqrt{84}$, since $-100$ is less than each of $\sqrt{84}$ and $-\sqrt{84}$. 
What you probably intend to say is that $b^2\lt 84$ iff $-\sqrt{84}\lt b\lt \sqrt{84}$. And we can rewrite this double inequality  as $|b|\lt\sqrt{84}$.
More simply, note that for any $b$, we have $\sqrt{b^2}=|b|$. To check this is true, verify it holds when $b\ge 0$ and when $b\lt 0$.
So we can conclude directly from $b^2\lt 84$ that $|b|\lt \sqrt{84}$.  And $\sqrt{84}=2\sqrt{21}$.
