What are the solutions to the equation y^4y+1=13 ?

The solutions to the equation y^4y+1=13 are y=\sqrt[5]{12}

Find the zeros of y^4y+1=13

The zeros of y^4y+1=13 are y=\sqrt[5]{12}

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y^{4} y + 1 = 13

Solution steps

y^{4} y + 1 = 13

Move

1

to the right side

y^{4} y = 12

Simplify

y^{5} = 12

For

x^{n} = f (a)

, n is odd, the solution is

2 x^{3} + 7 = 25

x^{3} + 2 x^{2} = 15 x

14 x (11 x - 2) (4 x + 13) = 0

(2 x - 1)^{3} = 0

5 x^{3} - 9 x^{2} + 3 x + 1 = 0

What are the solutions to the equation y^4y+1=13 ?
The solutions to the equation y^4y+1=13 are y=\sqrt[5]{12}
Find the zeros of y^4y+1=13
The zeros of y^4y+1=13 are y=\sqrt[5]{12}