What are the solutions to the equation 12y^2=-5y^2+2 ?

The solutions to the equation 12y^2=-5y^2+2 are y=sqrt(2/17),y=-sqrt(2/17)

Find the zeros of 12y^2=-5y^2+2

The zeros of 12y^2=-5y^2+2 are y=sqrt(2/17),y=-sqrt(2/17)

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12 y^{2} = - 5 y^{2} + 2

y = \frac{2}{17}, y = - \frac{2}{17}

Solution steps

12 y^{2} = - 5 y^{2} + 2

Move

5 y^{2}

to the left side

17 y^{2} = 2

Divide both sides by

17

y^{2} = \frac{2}{17}

For

x^{2} = f (a)

the solutions are

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

y = \frac{2}{17}, y = - \frac{2}{17}

(x - 4) (x - 6) = - 1

3 x^{2} + 9 x

x, 10 x^{2} - 8 y - 8 y^{3} = 0

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

(x - 1)^{2} + 7 = 56

What are the solutions to the equation 12y^2=-5y^2+2 ?
The solutions to the equation 12y^2=-5y^2+2 are y=sqrt(2/17),y=-sqrt(2/17)
Find the zeros of 12y^2=-5y^2+2
The zeros of 12y^2=-5y^2+2 are y=sqrt(2/17),y=-sqrt(2/17)