What are the solutions to the equation b^2=-1 ?

The solutions to the equation b^2=-1 are b=i,b=-i

Find the zeros of b^2=-1

The zeros of b^2=-1 are b=i,b=-i

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b^{2} = - 1

b = i, b = - i

Solution steps

b^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

b = - 1, b = - - 1

- 1 = i

- - 1 = - i

b = i, b = - i

y^{2} + (y + 5)^{2} - 5 = 16 (3 - y)

3244.8 = 3000 (1 + \frac{r}{1})^{2}

p, x^{2} - 2 p x + p^{2} + p - 6 = 0

3 x^{2} - 15 = 66

x^{2} - 6 x + 5 = - 3

What are the solutions to the equation b^2=-1 ?
The solutions to the equation b^2=-1 are b=i,b=-i
Find the zeros of b^2=-1
The zeros of b^2=-1 are b=i,b=-i