What are the solutions to the equation 2y^2+36=0 ?

The solutions to the equation 2y^2+36=0 are y=sqrt(2)*3i,y=-sqrt(2)*3i

Find the zeros of 2y^2+36=0

The zeros of 2y^2+36=0 are y=sqrt(2)*3i,y=-sqrt(2)*3i

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2 y^{2} + 36 = 0

y = 2 \cdot 3 i, y = - 2 \cdot 3 i

Solution steps

2 y^{2} + 36 = 0

Move

36

to the right side

2 y^{2} = - 36

Divide both sides by

2

y^{2} = - 18

For

x^{2} = f (a)

the solutions are

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

y = - 18, y = - - 18

- 18 = 2 \cdot 3 i

- - 18 = - 2 \cdot 3 i

y = 2 \cdot 3 i, y = - 2 \cdot 3 i

- 0.1 x^{2} + 0.1 x = - 0.06

2 x^{2} - 5 x + 2 = 0

5 x^{2} - 40 x + 5 = 0

5 x^{2} - 4 x - 6 = x^{2} - 7

5 x^{2} = 10 x + 15

What are the solutions to the equation 2y^2+36=0 ?
The solutions to the equation 2y^2+36=0 are y=sqrt(2)*3i,y=-sqrt(2)*3i
Find the zeros of 2y^2+36=0
The zeros of 2y^2+36=0 are y=sqrt(2)*3i,y=-sqrt(2)*3i