(x-7)^4-2(x-7)^2-15=0 equation

Given the equation:
$$\left(\left(x - 7\right)^{4} - 2 \left(x - 7\right)^{2}\right) - 15 = 0$$
Do replacement
$$v = \left(x - 7\right)^{2}$$
then the equation will be the:
$$v^{2} - 2 v - 15 = 0$$
This equation is of the form

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

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = -15$$
, then

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

(-2)^2 - 4 * (1) * (-15) = 64

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

v1 = (-b + sqrt(D)) / (2*a)

v2 = (-b - sqrt(D)) / (2*a)

or
$$v_{1} = 5$$
$$v_{2} = -3$$
The final answer:
Because
$$v = \left(x - 7\right)^{2}$$
then
$$x_{1} = \sqrt{v_{1}} + 7$$
$$x_{2} = 7 - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}} + 7$$
$$x_{4} = 7 - \sqrt{v_{2}}$$
then:
$$x_{1} = $$
$$\frac{5^{\frac{1}{2}}}{1} + \frac{7}{1} = \sqrt{5} + 7$$
$$x_{2} = $$
$$\frac{\left(-1\right) 5^{\frac{1}{2}}}{1} + \frac{7}{1} = 7 - \sqrt{5}$$
$$x_{3} = $$
$$\frac{7}{1} + \frac{\left(-3\right)^{\frac{1}{2}}}{1} = 7 + \sqrt{3} i$$
$$x_{4} = $$
$$\frac{7}{1} + \frac{\left(-1\right) \left(-3\right)^{\frac{1}{2}}}{1} = 7 - \sqrt{3} i$$