x^4+8x^2-9=0 equation

Given the equation:
$$x^{4} + 8 x^{2} - 9 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} + 8 v - 9 = 0$$
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
$$a = 1$$
$$b = 8$$
$$c = -9$$
, then

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

(8)^2 - 4 * (1) * (-9) = 100

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

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

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

or
$$v_{1} = 1$$
Simplify
$$v_{2} = -9$$
Simplify
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = \frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = \frac{0}{1} + \frac{1 \left(-9\right)^{\frac{1}{2}}}{1} = 3 i$$
$$x_{4} = \frac{0}{1} + \frac{\left(-1\right) \left(-9\right)^{\frac{1}{2}}}{1} = - 3 i$$