What are the solutions to the equation 2y^2=32 ?

The solutions to the equation 2y^2=32 are y=4,y=-4

Find the zeros of 2y^2=32

The zeros of 2y^2=32 are y=4,y=-4

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2 y^{2} = 32

y = 4, y = - 4

Solution steps

2 y^{2} = 32

Divide both sides by

2

y^{2} = 16

For

x^{2} = f (a)

the solutions are

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

y = 16, y = - 16

16 = 4

- 16 = - 4

y = 4, y = - 4

9 (2 x - 3)^{2} = 16

x^{2} - 19 x = 9.75

3 x^{2} - 13 x + 10 = 0

x^{2} - 19 x + 48 = 0

r, A = 2 π r (r + h)

What are the solutions to the equation 2y^2=32 ?
The solutions to the equation 2y^2=32 are y=4,y=-4
Find the zeros of 2y^2=32
The zeros of 2y^2=32 are y=4,y=-4