820_{/(1+h/6400)2} <10
thanks!!820(1+h6400)2<10
$\dfrac{820}{1+\left(\frac{h}{6400}\right)^2} < 10$
$1+\left(\frac{h}{6400}\right)^2 > 0$ so we can rearrange this as
$82 < 1+\left(\dfrac{h}{6400}\right)^2$
$81 < \left(\dfrac{h}{6400}\right)^2$
we'll assume that $h > 0$ so we have
$9 < \dfrac{h}{6400}$
$57600 < h$
Romsek had the square misplaced as DenisB and Plato point out. From
$\displaystyle \frac{820}{\left(1+ \frac{h}{6400}\right)^2}< 10$,
dividing both sides by 10 and multiplying both sides by $\displaystyle \left(1+ \frac{h}{6400}\right)^2$, both of which are positive, $\displaystyle 82< \left(1+ \frac{h}{6400}\right)^2$.
Taking the square root of both sides we have either
1) $\displaystyle -\sqrt{82}> 1+ \frac{h}{6400}$ or
2) $\displaystyle \sqrt{82}< 1+ \frac{h}{6400}$.
If (1), $\displaystyle \frac{h}{6400}< -1- \sqrt{82}$, so $\displaystyle h< -6400(1+ \sqrt{82})$.
If (2), $\displaystyle \frac{h}{6400}> \sqrt{82}- 1$, so $\displaystyle h> 6400(\sqrt{82}- 1)$.