t^2-t^4=0 equation

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

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

(1)^2 - 4 * (-1) * (0) = 1

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

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

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

or
$$v_{1} = 0$$
$$v_{2} = 1$$
The final answer:
Because
$$v = t^{2}$$
then
$$t_{1} = \sqrt{v_{1}}$$
$$t_{2} = - \sqrt{v_{1}}$$
$$t_{3} = \sqrt{v_{2}}$$
$$t_{4} = - \sqrt{v_{2}}$$
then:
$$t_{1} = $$
$$\frac{0^{\frac{1}{2}}}{1} + \frac{0}{1} = 0$$
$$t_{2} = $$
$$\frac{0}{1} + \frac{1^{\frac{1}{2}}}{1} = 1$$
$$t_{3} = $$
$$\frac{\left(-1\right) 1^{\frac{1}{2}}}{1} + \frac{0}{1} = -1$$