What are the solutions to the equation 16x^2+22=0 ?

The solutions to the equation 16x^2+22=0 are x=(sqrt(11))/(2sqrt(2))i,x=-(sqrt(11))/(2sqrt(2))i

Find the zeros of 16x^2+22=0

The zeros of 16x^2+22=0 are x=(sqrt(11))/(2sqrt(2))i,x=-(sqrt(11))/(2sqrt(2))i

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16 x^{2} + 22 = 0

x = \frac{11}{2 2} i, x = - \frac{11}{2 2} i

Solution steps

16 x^{2} + 22 = 0

Move

22

to the right side

16 x^{2} = - 22

Divide both sides by

16

x^{2} = - \frac{11}{8}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{11}{8}, x = - - \frac{11}{8}

- \frac{11}{8} = \frac{11}{2 2} i

- - \frac{11}{8} = - \frac{11}{2 2} i

x = \frac{11}{2 2} i, x = - \frac{11}{2 2} i

x^{2} + a x - b x = ab

3 x^{2} - 60 x + 741 = 0

10 + x (x - 2) = 2

8 v^{2} + 8 = - 32

17 x^{2} + 719 x = 0

What are the solutions to the equation 16x^2+22=0 ?
The solutions to the equation 16x^2+22=0 are x=(sqrt(11))/(2sqrt(2))i,x=-(sqrt(11))/(2sqrt(2))i
Find the zeros of 16x^2+22=0
The zeros of 16x^2+22=0 are x=(sqrt(11))/(2sqrt(2))i,x=-(sqrt(11))/(2sqrt(2))i