x^4+3x^2-4=0 equation

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

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

(3)^2 - 4 * (1) * (-4) = 25

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

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

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

or
$$v_{1} = 1$$
$$v_{2} = -4$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{2} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{\left(-4\right)^{\frac{1}{2}}}{1} = 2 i$$
$$x_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-4\right)^{\frac{1}{2}}}{1} = - 2 i$$