x^4−13x^2+36=0. equation

Given the equation:
$$x^{4} - 13 x^{2} + 36 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 13 v + 36 = 0$$
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 $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -13$$
$$c = 36$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 36 + \left(-13\right)^{2} = 25$$
Because D > 0, then the equation has two roots.
$$v_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$v_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$v_{1} = 9$$
Simplify
$$v_{2} = 4$$
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:
$$x_{1} = \frac{0}{1} + \frac{1 \cdot 9^{\frac{1}{2}}}{1} = 3$$
$$x_{2} = \frac{\left(-1\right) 9^{\frac{1}{2}}}{1} + \frac{0}{1} = -3$$
$$x_{3} = \frac{0}{1} + \frac{1 \cdot 4^{\frac{1}{2}}}{1} = 2$$
$$x_{4} = \frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$