What are the solutions to the equation 2t^2+1=0 ?

The solutions to the equation 2t^2+1=0 are t= 1/(sqrt(2))i,t=-1/(sqrt(2))i

Find the zeros of 2t^2+1=0

The zeros of 2t^2+1=0 are t= 1/(sqrt(2))i,t=-1/(sqrt(2))i

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2 t^{2} + 1 = 0

t = \frac{1}{2} i, t = - \frac{1}{2} i

Solution steps

2 t^{2} + 1 = 0

Move

1

to the right side

2 t^{2} = - 1

Divide both sides by

2

t^{2} = - \frac{1}{2}

For

x^{2} = f (a)

the solutions are

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

t = - \frac{1}{2}, t = - - \frac{1}{2}

- \frac{1}{2} = \frac{1}{2} i

- - \frac{1}{2} = - \frac{1}{2} i

t = \frac{1}{2} i, t = - \frac{1}{2} i

x^{3} + 2 x^{2} - 3 x

- 7 a = - 12 a + 65

25 k^{3} - k

1 - \frac{3}{11}

\frac{3}{2} + \frac{3}{2}

What are the solutions to the equation 2t^2+1=0 ?
The solutions to the equation 2t^2+1=0 are t= 1/(sqrt(2))i,t=-1/(sqrt(2))i
Find the zeros of 2t^2+1=0
The zeros of 2t^2+1=0 are t= 1/(sqrt(2))i,t=-1/(sqrt(2))i