What are the solutions to the equation 2y^2-32=0 ?

The solutions to the equation 2y^2-32=0 are y=4,y=-4

Find the zeros of 2y^2-32=0

The zeros of 2y^2-32=0 are y=4,y=-4

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2 y^{2} - 32 = 0

y = 4, y = - 4

Solution steps

2 y^{2} - 32 = 0

Move

32

to the right side

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

x, x^{2} + y^{2} - z^{2} = 1

7 x^{2} + 6 x - 1 = - 1

0.000006 x^{2} - 0.02 x + 25 = 60 - 0.025 x

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

3 w^{2} - 8 w + 4 = 0

What are the solutions to the equation 2y^2-32=0 ?
The solutions to the equation 2y^2-32=0 are y=4,y=-4
Find the zeros of 2y^2-32=0
The zeros of 2y^2-32=0 are y=4,y=-4