(2x^2-5-7)(x-1)=0 equation

Given the equation:
$$\left(x - 1\right) \left(\left(2 x^{2} - 5\right) - 7\right) = 0$$
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 1 = 0$$
$$2 x^{2} - 12 = 0$$
solve the resulting equation:
1.
$$x - 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 1$$
We get the answer: x1 = 1
2.
$$2 x^{2} - 12 = 0$$
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:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 2$$
$$b = 0$$
$$c = -12$$
, then

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

(0)^2 - 4 * (2) * (-12) = 96

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

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

x3 = (-b - sqrt(D)) / (2*a)

or
$$x_{2} = \sqrt{6}$$
$$x_{3} = - \sqrt{6}$$
The final answer:
$$x_{1} = 1$$
$$x_{2} = \sqrt{6}$$
$$x_{3} = - \sqrt{6}$$