x=√(4x-3) equation

Given the equation
$$x = \sqrt{4 x - 3}$$
Transfer the right side of the equation left part with negative sign
$$- \sqrt{4 x - 3} = - x$$
We raise the equation sides to 2-th degree
$$4 x - 3 = x^{2}$$
$$4 x - 3 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 4 x - 3 = 0$$
This equation is of the form

a*x^2 + b*x + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 4$$
$$c = -3$$
, then

D = b^2 - 4 * a * c = 

(4)^2 - 4 * (-1) * (-3) = 4

Because D > 0, then the equation has two roots.

x1 = (-b + sqrt(D)) / (2*a)

x2 = (-b - sqrt(D)) / (2*a)

or
$$x_{1} = 1$$
$$x_{2} = 3$$

Because
$$\sqrt{4 x - 3} = x$$
and
$$\sqrt{4 x - 3} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{1} = 1$$
$$x_{2} = 3$$