What are the solutions to the equation -3y^2+12=0 ?

The solutions to the equation -3y^2+12=0 are y=2,y=-2

Find the zeros of -3y^2+12=0

The zeros of -3y^2+12=0 are y=2,y=-2

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- 3 y^{2} + 12 = 0

y = 2, y = - 2

Solution steps

- 3 y^{2} + 12 = 0

Move

12

to the right side

- 3 y^{2} = - 12

Divide both sides by

- 3

y^{2} = 4

For

x^{2} = f (a)

the solutions are

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

y = 4, y = - 4

4 = 2

- 4 = - 2

y = 2, y = - 2

7 x^{2} + 8 x - 2 = 0

(20 + x) (35 + x) = 1270

2 x (5 x - 1) = 9

10 x^{2} - 16 x - 8 = 0

0.6 x^{2} + 6 x + 2 = 0

What are the solutions to the equation -3y^2+12=0 ?
The solutions to the equation -3y^2+12=0 are y=2,y=-2
Find the zeros of -3y^2+12=0
The zeros of -3y^2+12=0 are y=2,y=-2