For the first one, \(\displaystyle \frac{8}{\sqrt{60}\), I assume you want to simplify it?
Well, \(\displaystyle \sqrt{60} = \sqrt{4\cdot15} = \sqrt{4}\cdot\sqrt{15} = 2\sqrt{15}\), so you end up with \(\displaystyle \frac{4}{\sqrt{15}}\)
Note that the radical, \(\displaystyle \sqrt{blah}\) denotes to the positive square root (of "blah").
For the second one, you have written the question fine (well, it's unambiguous, anyway).
-3(4x - 9)^2 + 5 = -3
Move that 5 to the right-hand side by subtracting both sides by 5:
-3(4x - 9)^2 = -8
Divide both sides by -3:
\(\displaystyle (4x - 9)^2 = \frac{8}{3}\)
Take square roots of both sides:
\(\displaystyle 4x - 9 = \pm \sqrt{\frac{8}{3}}\)
Move the -9 to the right-hand side:
\(\displaystyle 4x = 9 \pm \sqrt{\frac{8}{3}}\)
Divide both sides by 4:
\(\displaystyle x = \frac{9 \pm \sqrt{\frac{8}{3}}}{4}\)
We can pay around with the radicals:
\(\displaystyle \sqrt{\frac{8}{3}} = \frac{\sqrt{8}}{\sqrt{3}}\)
\(\displaystyle = \frac{\sqrt{4\cdot2}}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}}\)
And you end up with:
\(\displaystyle x = \frac{9}{4} \pm \frac{\sqrt{2}}{2\sqrt{3}}\)
Your version of the provided answer, I think, translates to:
\(\displaystyle \frac{9 \pm 2\sqrt{3}}{4}\)
... so perhaps something was lost in translation?
Edit: Looking at my answer, I think it would be best to rationlise the denominator:
\(\displaystyle \frac{\sqrt{2}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{6}}{6}\)
