(x+3)^4-4*(x+3)^2-5=0 equation

Given the equation:
$$\left(\left(x + 3\right)^{4} - 4 \left(x + 3\right)^{2}\right) - 5 = 0$$
Do replacement
$$v = \left(x + 3\right)^{2}$$
then the equation will be the:
$$v^{2} - 4 v - 5 = 0$$
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
$$a = 1$$
$$b = -4$$
$$c = -5$$
, then

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

(-4)^2 - 4 * (1) * (-5) = 36

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

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

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

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