What are the solutions to the equation solvefor x,z^3+3z+y=x^2+1 ?

The solutions to the equation solvefor x,z^3+3z+y=x^2+1 are x=sqrt(z^3+3z+y-1),x=-sqrt(z^3+3z+y-1)

Find the zeros of solvefor x,z^3+3z+y=x^2+1

The zeros of solvefor x,z^3+3z+y=x^2+1 are x=sqrt(z^3+3z+y-1),x=-sqrt(z^3+3z+y-1)

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solve for

x, z^{3} + 3 z + y = x^{2} + 1

x = z^{3} + 3 z + y - 1, x = - z^{3} + 3 z + y - 1

Solution steps

z^{3} + 3 z + y = x^{2} + 1

Switch sides

x^{2} + 1 = z^{3} + 3 z + y

Move

1

to the right side

x^{2} = z^{3} + 3 z + y - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = z^{3} + 3 z + y - 1, x = - z^{3} + 3 z + y - 1

x^{2} - 4 x + 9

x^{2} - 4 x - 3

p^{2} = 169

1.5 x^{2} - 10 x - 16 = 0

x^{2} - 23 x - 50 = 0

What are the solutions to the equation solvefor x,z^3+3z+y=x^2+1 ?
The solutions to the equation solvefor x,z^3+3z+y=x^2+1 are x=sqrt(z^3+3z+y-1),x=-sqrt(z^3+3z+y-1)
Find the zeros of solvefor x,z^3+3z+y=x^2+1
The zeros of solvefor x,z^3+3z+y=x^2+1 are x=sqrt(z^3+3z+y-1),x=-sqrt(z^3+3z+y-1)