Help with quadratic equation and square roots 3x^4-35x^2+72=0

I know how to get all the way to the end of the problem, but can't figure out how to get the answer. I know what the answer is, but need help.
Equation:
3x^4-35x^2+72=0
3*y^2 - 35*y + 72 = 0.
(35 +/- sqrt(35^2 - 4*3*72)) / (2*3)
y = (35 +/- sqrt(361)) / 6 = (35 +/- 19) / 6 = 9 or 16/6=8/3.

Answer: +/- 3, +/- 2*sqrt(6) / 3.

My question is how do I convert 8/3 into +/- 2√6/3

Also, how do I solve this problem? It's a practice question without an example.
3x^2=5
Thanks!

You forgot the last step of the problem. You only solved for the solution of y. The question is asking you for x.
x^2 = y
x = ±√y
Substitute each solution for y to solve for x.
x = ±√9 = ±3
x = ±√(8/3)
= ±√8 / √3
= ±2√2 / √3
= ±2√2*√3 / √3*√3
= ±2√2*3 / √3^2
= ±2√6 / 3

The other problem.
3x^2 = 5
x^2 = 5/3
x = ±√(5/3)
x = ±√5 / √3
x = ±√5*√3 / √3*√3
x = ±√5*3 / √3^2
x = ±√15 / 3

3x^4 - 35x^2 + 72 = 0
let x^2 = y
3y^2 - 35y + 72 = 0
(y-9)(3y-8) = 0
y-9 = 0
y = 9
x^2 = 9
x = +/-3

3y-8 = 0
3y = 8
3x^2 = 8
x^2 = 8/3
x = +/-sqrt(8/3) = +/- sqrt(24/9) =
+/- (2/3)* sqrt6 or +/- 2sqrt(6) /3

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3x^2 = 5
x^2 = 5/3
x = +/- sqrt(5/3)
you don't want to have a radical in the denominator, so multiply by 3/3

x = +/- sqrt(15/9) = +/- sqrt(15) /3