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

The solutions to the equation 2x^2+2=0 are x=i,x=-i

Find the zeros of 2x^2+2=0

The zeros of 2x^2+2=0 are x=i,x=-i

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2 x^{2} + 2 = 0

x = i, x = - i

Solution steps

2 x^{2} + 2 = 0

Move

2

to the right side

2 x^{2} = - 2

Divide both sides by

2

x^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

x = - 1, x = - - 1

- 1 = i

- - 1 = - i

x = i, x = - i

11 - 62

\frac{2}{5} (x - 16) = - 6

(4 n^{4} - 8 n + 4) - (8 n^{2} + 4 n^{4} + 1)

(x - 2)^{2} = 45

a c + 7 a - 2 c - 14

What are the solutions to the equation 2x^2+2=0 ?
The solutions to the equation 2x^2+2=0 are x=i,x=-i
Find the zeros of 2x^2+2=0
The zeros of 2x^2+2=0 are x=i,x=-i