What are the solutions to the equation 6y^4=7-3y^2 ?

The solutions to the equation 6y^4=7-3y^2 are

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6y^4=7-3y^2

6 y^{4} = 7 - 3 y^{2}

y = \frac{- 3 + 177}{2 3}, y = - \frac{- 3 + 177}{2 3}, y = \frac{- 3 - 177}{12}, y = - \frac{- 3 - 177}{12}

Solution steps

6 y^{4} = 7 - 3 y^{2}

Move

3 y^{2}

to the left side

6 y^{4} + 3 y^{2} = 7

Move

7

to the left side

6 y^{4} + 3 y^{2} - 7 = 0

Rewrite the equation with

u = y^{2}

and

u^{2} = y^{4}

6 u^{2} + 3 u - 7 = 0

Solve

6 u^{2} + 3 u - 7 = 0 : u = \frac{- 3 + 177}{12}, u = \frac{- 3 - 177}{12}

u = \frac{- 3 + 177}{12}, u = \frac{- 3 - 177}{12}

Substitute back

u = y^{2},

solve for

y

Solve

y^{2} = \frac{- 3 + 177}{12} : y = \frac{- 3 + 177}{2 3}, y = - \frac{- 3 + 177}{2 3}

Solve

y^{2} = \frac{- 3 - 177}{12} : y = \frac{- 3 - 177}{12}, y = - \frac{- 3 - 177}{12}

The solutions are

y = \frac{- 3 + 177}{2 3}, y = - \frac{- 3 + 177}{2 3}, y = \frac{- 3 - 177}{12}, y = - \frac{- 3 - 177}{12}

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

4 \cdot (x^{4} + 1) = 5 x^{2}

r^{3} - 3 r^{2} + 2 r - 1 = 0

x^{2} (4 x + 5) - (4 x + 5) = 0

x^{3} - 11 x^{2} - 11 x + 121 = 0

What are the solutions to the equation 6y^4=7-3y^2 ?
The solutions to the equation 6y^4=7-3y^2 are
Find the zeros of 6y^4=7-3y^2
The zeros of 6y^4=7-3y^2 are