You need to raise to square to remove the radical to the left side, such that:

`(sqrt(3x - 4))^2 = x^2 => 3x - 4 = x^2`

`x^2 - 3x + 4 = 0 => x^2 - 3x = - 4 `

You need to complete the square to the left side, such that:

`x^2 - 3x + 9/4 = 9/4 - 4 => (x - 3/2)^2 = (9 - 16)/4`

`(x - 3/2)^2 = (-7)/4 `

**Hence, `(x - 3/2)^2 > 0 ` for `x in (-oo,oo)` , thus, there exists no real solution to the given equation.**

The value of x has to be determined given that (3x-4)^1/2=x

(3x-4)^1/2=x

take the square of both the sides

3x - 4 = x^2

x^2 - 3x + 4 = 0

Solve this quadratic equation x^2 - 3x + 4 = 0

In the equation x^2 - 3x + 4 = 0, (-3)^2 - 4*1*4 = 9 - 16 = - 7

For a quadratic equation ax^2 + bx + c = 0 if b^2 < 4ac the equation does not have any real roots. This is the case with the quadratic equation that has been derived.

As the equation has no real roots, the original equation (3x-4)^1/2=x also does not have any real solutions.