What are the solutions to the equation 2x^2+40=0 ?

The solutions to the equation 2x^2+40=0 are x=2sqrt(5)i,x=-2sqrt(5)i

Find the zeros of 2x^2+40=0

The zeros of 2x^2+40=0 are x=2sqrt(5)i,x=-2sqrt(5)i

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2 x^{2} + 40 = 0

x = 25 i, x = - 25 i

Solution steps

2 x^{2} + 40 = 0

Move

40

to the right side

2 x^{2} = - 40

Divide both sides by

2

x^{2} = - 20

For

x^{2} = f (a)

the solutions are

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

x = - 20, x = - - 20

- 20 = 25 i

- - 20 = - 25 i

x = 25 i, x = - 25 i

6 x^{2} + 5 x - 50

(x + 2 - 3 i) (x + 2 + 3 i)

\frac{2 - x}{4} < \frac{4 x - 8}{9}

9 a^{2} - 25 b^{2}

(3 x^{2} y^{2})^{3}

What are the solutions to the equation 2x^2+40=0 ?
The solutions to the equation 2x^2+40=0 are x=2sqrt(5)i,x=-2sqrt(5)i
Find the zeros of 2x^2+40=0
The zeros of 2x^2+40=0 are x=2sqrt(5)i,x=-2sqrt(5)i