What are the solutions to the equation 2x^2+9=-41 ?

The solutions to the equation 2x^2+9=-41 are x=5i,x=-5i

Find the zeros of 2x^2+9=-41

The zeros of 2x^2+9=-41 are x=5i,x=-5i

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2 x^{2} + 9 = - 41

x = 5 i, x = - 5 i

Solution steps

2 x^{2} + 9 = - 41

Move

9

to the right side

2 x^{2} = - 50

Divide both sides by

2

x^{2} = - 25

For

x^{2} = f (a)

the solutions are

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

x = - 25, x = - - 25

- 25 = 5 i

- - 25 = - 5 i

x = 5 i, x = - 5 i

2 (- 5)

\frac{270}{180} π

x^{2} + (\frac{3}{2} x - \frac{1}{4}) = 0

3 (\frac{2}{3})^{3}

\frac{2 2}{3 - 2}

What are the solutions to the equation 2x^2+9=-41 ?
The solutions to the equation 2x^2+9=-41 are x=5i,x=-5i
Find the zeros of 2x^2+9=-41
The zeros of 2x^2+9=-41 are x=5i,x=-5i