What are the solutions to the equation 2x^2+31=9 ?

The solutions to the equation 2x^2+31=9 are x=sqrt(11)i,x=-sqrt(11)i

Find the zeros of 2x^2+31=9

The zeros of 2x^2+31=9 are x=sqrt(11)i,x=-sqrt(11)i

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2 x^{2} + 31 = 9

x = 11 i, x = - 11 i

Solution steps

2 x^{2} + 31 = 9

Move

31

to the right side

2 x^{2} = - 22

Divide both sides by

2

x^{2} = - 11

For

x^{2} = f (a)

the solutions are

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

x = - 11, x = - - 11

- 11 = 11 i

- - 11 = - 11 i

x = 11 i, x = - 11 i

π + \frac{π}{4}

2 (2 x - 7) = 22

\frac{1}{\frac{π}{4}}

2 (2 - 3 x) - 3 (3 - 2 x) = 4 (x + 1) + 3 (4 - 5 x)

x^{4} + 2 x^{3} - 5 x^{2} - 12 x - 4

What are the solutions to the equation 2x^2+31=9 ?
The solutions to the equation 2x^2+31=9 are x=sqrt(11)i,x=-sqrt(11)i
Find the zeros of 2x^2+31=9
The zeros of 2x^2+31=9 are x=sqrt(11)i,x=-sqrt(11)i