(x+3)(x+1)(x-7)=(x+3)(x+1)(x-8) equation

Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$\left(x + 1\right) \left(x + 3\right) \left(x - 7\right) = \left(x + 1\right) \left(x + 3\right) \left(x - 8\right)$$
to
$$- \left(x + 1\right) \left(x + 3\right) \left(x - 8\right) + \left(x + 1\right) \left(x + 3\right) \left(x - 7\right) = 0$$
Expand the expression in the equation
$$- \left(x + 1\right) \left(x + 3\right) \left(x - 8\right) + \left(x + 1\right) \left(x + 3\right) \left(x - 7\right) = 0$$
We get the quadratic equation
$$x^{2} + 4 x + 3 = 0$$
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
$$x_{1} = -1$$
$$x_{2} = -3$$