What are the solutions to the equation -3(x+12)=x^6-(1-x) ?

The solutions to the equation -3(x+12)=x^6-(1-x) are No Solution for x\in\mathbb{R}

Find the zeros of -3(x+12)=x^6-(1-x)

The zeros of -3(x+12)=x^6-(1-x) are No Solution for x\in\mathbb{R}

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- 3 (x + 12) = x^{6} - (1 - x)

No Solution for x \in R

Solution steps

- 3 (x + 12) = x^{6} - (1 - x)

Expand

- 3 (x + 12) : - 3 x - 36

Expand

x^{6} - (1 - x) : x^{6} - 1 + x

- 3 x - 36 = x^{6} - 1 + x

Switch sides

x^{6} - 1 + x = - 3 x - 36

Move

36

to the left side

x^{6} + x + 35 = - 3 x

Move

3 x

to the left side

x^{6} + 4 x + 35 = 0

Find one solution for

x^{6} + 4 x + 35 = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

No Solution for x \in R

x^{2} + 5 x - 84

\frac{2 x ^{2} - 2 x - 4}{x ^{2} - 4} > 0

90 x^{2} - 200 y^{2}

- 3 x^{2} + 5 x + 3 = 0

2 y^{2} - 11 y + 5

What are the solutions to the equation -3(x+12)=x^6-(1-x) ?
The solutions to the equation -3(x+12)=x^6-(1-x) are No Solution for x\in\mathbb{R}
Find the zeros of -3(x+12)=x^6-(1-x)
The zeros of -3(x+12)=x^6-(1-x) are No Solution for x\in\mathbb{R}