x^4-17x^2+16=0 equation

Given the equation:
$$\left(x^{4} - 17 x^{2}\right) + 16 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 17 v + 16 = 0$$
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
$$a = 1$$
$$b = -17$$
$$c = 16$$
, then

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

(-17)^2 - 4 * (1) * (16) = 225

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

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

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

or
$$v_{1} = 16$$
$$v_{2} = 1$$
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:
$$x_{1} = $$
$$\frac{0}{1} + \frac{16^{\frac{1}{2}}}{1} = 4$$
$$x_{2} = $$
$$\frac{\left(-1\right) 16^{\frac{1}{2}}}{1} + \frac{0}{1} = -4$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$x_{4} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$