What are the solutions to the equation t=-3.9(t)^2-(t)^4 ?

The solutions to the equation t=-3.9(t)^2-(t)^4 are t=0,t\approx-0.25229…

Find the zeros of t=-3.9(t)^2-(t)^4

The zeros of t=-3.9(t)^2-(t)^4 are t=0,t\approx-0.25229…

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t = - 3.9 (t)^{2} - (t)^{4}

t = 0, t \approx - 0.25229 \dots

Solution steps

t = - 3.9 t^{2} - t^{4}

Switch sides

- 3.9 t^{2} - t^{4} = t

Move

t

to the left side

- 3.9 t^{2} - t^{4} - t = 0

Factor

- 3.9 t^{2} - t^{4} - t : - t (t^{3} + 3.9 t + 1)

- t (t^{3} + 3.9 t + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

t = 0 or t^{3} + 3.9 t + 1 = 0

Solve

t^{3} + 3.9 t + 1 = 0 : t \approx - 0.25229 \dots

The solutions are

t = 0, t \approx - 0.25229 \dots

x, y = q^{2} + (a x)^{3}

x^{4} = 24 x + 32

12 (k^{3} - 4 k^{2} + k + 6) = 0

q^{3} - 10 q^{2} - 5808 = 0

1.05 = (1 + \frac{x}{12})^{12}

What are the solutions to the equation t=-3.9(t)^2-(t)^4 ?
The solutions to the equation t=-3.9(t)^2-(t)^4 are t=0,t\approx-0.25229…
Find the zeros of t=-3.9(t)^2-(t)^4
The zeros of t=-3.9(t)^2-(t)^4 are t=0,t\approx-0.25229…