What are the solutions to the equation solvefor y,4y^2+18=96x ?

The solutions to the equation solvefor y,4y^2+18=96x are y=sqrt((3(16x-3))/2),y=-sqrt((3(16x-3))/2)

Find the zeros of solvefor y,4y^2+18=96x

The zeros of solvefor y,4y^2+18=96x are y=sqrt((3(16x-3))/2),y=-sqrt((3(16x-3))/2)

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solve for

y, 4 y^{2} + 18 = 96 x

y = \frac{3 ( 16 x - 3 )}{2}, y = - \frac{3 ( 16 x - 3 )}{2}

Solution steps

4 y^{2} + 18 = 96 x

Move

18

to the right side

4 y^{2} = 96 x - 18

Divide both sides by

4

y^{2} = \frac{3 ( 16 x - 3 )}{2}

For

x^{2} = f (a)

the solutions are

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

y = \frac{3 ( 16 x - 3 )}{2}, y = - \frac{3 ( 16 x - 3 )}{2}

2 q^{2} + 9 = - q

(x + 3) (x + 7) = 11

2 c^{2} - 3 c - 9 = 0

(3 x + 5)^{2} - \frac{1}{3} = 0

X^{2} - 4 X + 1 = 0

What are the solutions to the equation solvefor y,4y^2+18=96x ?
The solutions to the equation solvefor y,4y^2+18=96x are y=sqrt((3(16x-3))/2),y=-sqrt((3(16x-3))/2)
Find the zeros of solvefor y,4y^2+18=96x
The zeros of solvefor y,4y^2+18=96x are y=sqrt((3(16x-3))/2),y=-sqrt((3(16x-3))/2)