What are the solutions to the equation 2-3-2x^4-3x-16+2x-3=x+12,a ?

The solutions to the equation 2-3-2x^4-3x-16+2x-3=x+12,a are No Solution for x\in\mathbb{R}

Find the zeros of 2-3-2x^4-3x-16+2x-3=x+12,a

The zeros of 2-3-2x^4-3x-16+2x-3=x+12,a are No Solution for x\in\mathbb{R}

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2 - 3 - 2 x^{4} - 3 x - 16 + 2 x - 3 = x + 12, a

No Solution for x \in R

Solution steps

2 - 3 - 2 x^{4} - 3 x - 16 + 2 x - 3 = x + 12

Refine

- 2 x^{4} - x - 20 = x + 12

Move

12

to the left side

- 2 x^{4} - x - 32 = x

Move

x

to the left side

- 2 x^{4} - 2 x - 32 = 0

Find one solution for

- 2 x^{4} - 2 x - 32 = 0

using Newton-Raphson:

No Solution for

x \in R

The solution is

No Solution for x \in R

- x^{5} - 1 = 0

y, 5 (y + 1)^{3} + x^{2} = 1

0 = 24 x^{3} + 2 x^{2} + x - 5

25 x^{4} - 65 x^{2} + 36 = 0

24 = r^{4}

What are the solutions to the equation 2-3-2x^4-3x-16+2x-3=x+12,a ?
The solutions to the equation 2-3-2x^4-3x-16+2x-3=x+12,a are No Solution for x\in\mathbb{R}
Find the zeros of 2-3-2x^4-3x-16+2x-3=x+12,a
The zeros of 2-3-2x^4-3x-16+2x-3=x+12,a are No Solution for x\in\mathbb{R}