What are the solutions to the equation w^2+32=0 ?

The solutions to the equation w^2+32=0 are w=4sqrt(2)i,w=-4sqrt(2)i

Find the zeros of w^2+32=0

The zeros of w^2+32=0 are w=4sqrt(2)i,w=-4sqrt(2)i

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w^{2} + 32 = 0

w = 42 i, w = - 42 i

Solution steps

w^{2} + 32 = 0

Move

32

to the right side

w^{2} = - 32

For

x^{2} = f (a)

the solutions are

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

w = - 32, w = - - 32

- 32 = 42 i

- - 32 = - 42 i

w = 42 i, w = - 42 i

\frac{1}{5} (q + 3) = \frac{1}{2} (q - 6)

5.1 - r \geq 5

x^{2} + 12 x + 33 = 0

x > - 1 and x \leq 2

32 x^{4} y^{9}

What are the solutions to the equation w^2+32=0 ?
The solutions to the equation w^2+32=0 are w=4sqrt(2)i,w=-4sqrt(2)i
Find the zeros of w^2+32=0
The zeros of w^2+32=0 are w=4sqrt(2)i,w=-4sqrt(2)i