What are the solutions to the equation -2=2x^2+8 ?

The solutions to the equation -2=2x^2+8 are x=sqrt(5)i,x=-sqrt(5)i

Find the zeros of -2=2x^2+8

The zeros of -2=2x^2+8 are x=sqrt(5)i,x=-sqrt(5)i

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- 2 = 2 x^{2} + 8

x = 5 i, x = - 5 i

Solution steps

- 2 = 2 x^{2} + 8

Switch sides

2 x^{2} + 8 = - 2

Move

8

to the right side

2 x^{2} = - 10

Divide both sides by

2

x^{2} = - 5

For

x^{2} = f (a)

the solutions are

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

x = - 5, x = - - 5

- 5 = 5 i

- - 5 = - 5 i

x = 5 i, x = - 5 i

2 k^{2} + 11 k + 15

\frac{( 15 - 8 i ) + ( 12 - 6 i )}{( - 14 + 3 i ) - ( - 19 - 4 i )}

\frac{1}{4} x^{2} - 2 x + 3 \leq 0

0 = \frac{- 32}{81 ( x ^{2} - x - 20 )}

What are the solutions to the equation -2=2x^2+8 ?
The solutions to the equation -2=2x^2+8 are x=sqrt(5)i,x=-sqrt(5)i
Find the zeros of -2=2x^2+8
The zeros of -2=2x^2+8 are x=sqrt(5)i,x=-sqrt(5)i