I need help with math :(?

the steps to solve?

2x^(2/3) - 5x^(1/3) - 3 = 0 => for principal roots the domain is x ≥ 0, then:

let u = x^(/13) , then: x^(2/3) = u^2, then the hidden quadratic can be seen as:

2u^2 - 5u - 3 = 0

2u^2 - 6u + u - 3 = 0

2u(u - 3) + (u - 3) = 0

(2u + 1)(u - 3) = 0

u = -1/2 , 3

x^(1/3) = -1/2 => reject since using the principal i.e the positive root only:

x^(1/3) = 3

x = 3^3 = 27 => only principal solution

Edit: I'm editing my solution above and adding more below to incorporate the following correction and please accept my sincere apologies:

2x^(2/3) - 5x^(1/3) - 3 = 0, for real value roots the domain is all the real numbers, then:
u = -1/2 , 3
x^(1/3) = -1/2 => x = -1/8
x^(1/3) = 3 => x = 27
so x = -1/8 , 27 => both are real valued roots.

Regards.

Lets use a substitution , let y = x^(1/3)

We now have :

2y^2 - 5y - 3 = 0

This is solved like a typical quadratic by factorising into two brackets :

2y^2 - 5y - 3 = 0

(2y +1)(y-3) = 0

2y + 1 = 0
y = -1/2

or y-3 = 0
y = 3

Replacing y with our x^(1/3) gives us:

x^1/3 = -1/2

x = -1/8

or x^1/3 = 3

x = 27

So the answers are -1/8 and 27

let a=x^(1/3) then you have:

2a^2-5a-3=0

2a^2+a-6a-3=0

a(2a+1)-3(2a+1)=0

(2a+1)(a-3)=0 and since a=x^(1/3)...

(2x^(1/3)+1)(x^(1/3)-3)=0

x=(-1/8, 27)

Put u = x^(1/3). Then 2u^2 - 5u - 3 = 0, ie.,
(2u+1)(u-3) = 0 & u = (-1/2)..(i) or u = 3..(ii). Now,
for (i) holding, x^(1/3) = -(1/2) and x = [-(1/2)]^3
= - (1/8). For (ii) holding, x^(1/3) = 3 and x = 27.
Therefore x = - (1/8) or 27.

Please read :

x^(1/3) = t

2t^2 - 5t - 3 = 0
t = (5 +/- sqrt(25 + 24)) / 4
t = (5 +/- 7) / 4
t = 12/4 , -2/4
t = 3 , -1/2
x^(1/3) = 3 , -1/2
x = 3^3 , (-1/2)^3
x = 8 , -1/8