Nah, post in the thread if it's something to do with the question. PM (which is private message) is if you wanted to chat about something more general or get in touch with me about something other than a question. Questions go in threads like this one.

[tex]\pm[/tex] is $\displaystyle \pm$ and [tex]\sqrt{ax+b}[/tex] is $\displaystyle \sqrt{ax+b}$. You'll pick up latex tricks as you post

I'm not sure what happened to your 10 on the denominator, that should still be there on the second term. In the numerator you can factor out 20 (which I shall write as 4*5)

$\displaystyle h = -1 \pm \dfrac{\sqrt{4 \times 5(d^2-5)}}{10} = -1 \pm \dfrac{2 \sqrt{5(d^2-5)}}{10}$

Here is where the hint I suggested comes in. Write that 5 under the radical as $\displaystyle \dfrac{25}{5}$ which is $\displaystyle h = -1 \pm \dfrac{2 \sqrt{\dfrac{25}{5}(d^2-5)}}{10}$

As $\displaystyle 25 =5^2$ we can pull that 25 out of the radical: $\displaystyle h = -1 \pm \dfrac{2 \times 5 \sqrt{\dfrac{d^2-5}{5}}}{10} = -1 \pm \sqrt{\dfrac{d^2-5}{5}$

QED