How to solve $ 2x-3-2x^{ -1/2 }= 0 $ for $x$? Equation to solve
$$
2x-3-2x^{ -(1/2) }= 0
$$
The answer should be $2.1777$. However I'm not too sure how the steps in between are constructed. Anyone can guide me how do I solve x for this equation?
Progress
One way that I have tried solving this is
$$
2x-3-2x^{ -(1/2) }= 0
$$
$$
2x-2x^{ -(1/2) }= 3
$$
$$
2(x-x^{ -(1/2) })= 3
$$
$$
x-x^{ -(1/2) }= \dfrac32
$$
Not very sure how do I proceed on from here and whether it is the correct way of doing it?
 A: $$2x-3-\frac2{\sqrt x}=0\implies(2x-3)^2=\left(\frac2{\sqrt x}\right)^2\implies$$
$$4x^2-12x+9=\frac4x\implies 4x^3-12x^2+9x-4=0$$
I don't think the above has a simple/elementary solution. Are you sure you wrote the correct equation?
A: The equation is not a quadratic equation. If we let $x^{1/2} = u$, we then get that
$$2u^2 - 3 - \dfrac2u = 0 \implies u^3 - \dfrac32u - 1 = 0$$
This is a depressed cubic and solving such depressed cubic can be found here. The solution is given as follows:
Let $\alpha = \dfrac{\sqrt[3]{108-54 \sqrt2}}6$, $\beta = \dfrac{\sqrt[3]{2+\sqrt2}}{2^{2/3}}$ and $\omega = \dfrac{1+i \sqrt3}2$; then the three solutions are
$$\alpha + \beta; -\alpha \omega - \beta \bar{\omega}; -\alpha \bar{\omega} - \beta \omega$$ Of the three the only real solution is $\alpha + \beta$. Since, $x^{1/2} = u$, the only real solution for the initial equation is $(\alpha + \beta)^2$, which evaluates to $2.177650698804059954962643867932125475959166$ approximately. 
A: Let's suppose, just for grins, that the equation is this slight variation:
$$2x^{1/2} -3 -2x^{-1/2}=0$$
Then multiplying through by $x^{1/2}$ yields $2x -3x^{1/2} -2 = 0$
$$ \cdots = (2x^{1/2} +1)(x^{1/2} -2) = 0$$
