sqrt(2x+3)+sqrt(x-2)=2*sqrt(x-1) equation

Given the equation
$$\sqrt{x - 2} + \sqrt{2 x + 3} = 2 \sqrt{x - 1}$$
We raise the equation sides to 2-th degree
$$\left(\sqrt{x - 2} + \sqrt{2 x + 3}\right)^{2} = 4 x - 4$$
or
$$\left(-2\right)^{2} \left(x - 1\right) + \left(\left(-2\right) 2 \sqrt{\left(x - 2\right) \left(x - 1\right)} + 1^{2} \left(x - 2\right)\right) = 4 x - 4$$
or
$$5 x - 4 \sqrt{x^{2} - 3 x + 2} - 6 = 4 x - 4$$
transform:
$$- 4 \sqrt{x^{2} - 3 x + 2} = 2 - x$$
We raise the equation sides to 2-th degree
$$16 x^{2} - 48 x + 32 = \left(2 - x\right)^{2}$$
$$16 x^{2} - 48 x + 32 = x^{2} - 4 x + 4$$
Transfer the right side of the equation left part with negative sign
$$15 x^{2} - 44 x + 28 = 0$$
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
$$a = 15$$
$$b = -44$$
$$c = 28$$
, then

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

(-44)^2 - 4 * (15) * (28) = 256

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

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

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

or
$$x_{1} = 2$$
$$x_{2} = \frac{14}{15}$$

Because
$$\sqrt{x^{2} - 3 x + 2} = \frac{x}{4} - \frac{1}{2}$$
and
$$\sqrt{x^{2} - 3 x + 2} \geq 0$$
then
$$\frac{x}{4} - \frac{1}{2} \geq 0$$
or
$$2 \leq x$$
$$x < \infty$$
$$x_{1} = 2$$
check:
$$x_{1} = 2$$
$$\sqrt{x_{1} - 2} - 2 \sqrt{x_{1} - 1} + \sqrt{2 x_{1} + 3} = 0$$
=
$$- 2 \sqrt{-1 + 2} + \left(\sqrt{-2 + 2} + \sqrt{3 + 2 \cdot 2}\right) = 0$$
=

-2 + sqrt(7) = 0

- No
The final answer:
This equation has no roots