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

The solutions to the equation 2x^2+32=0 are x=4i,x=-4i

Find the zeros of 2x^2+32=0

The zeros of 2x^2+32=0 are x=4i,x=-4i

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2 x^{2} + 32 = 0

x = 4 i, x = - 4 i

Solution steps

2 x^{2} + 32 = 0

Move

32

to the right side

2 x^{2} = - 32

Divide both sides by

2

x^{2} = - 16

For

x^{2} = f (a)

the solutions are

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

x = - 16, x = - - 16

- 16 = 4 i

- - 16 = - 4 i

x = 4 i, x = - 4 i

x - \frac{3 x - 2}{10} = \frac{16 x - 17}{20} + 1

\frac{( 25 ) \cdot 60}{10}

7 - 13

- 5 \leq 3 - 2 x < 8

- 2.7 (920 - 2 t) \cdot (920 - 2 t)^{- \frac{7}{2}}

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