What are the solutions to the equation 8a^2-4=-5 ?

The solutions to the equation 8a^2-4=-5 are a= 1/(2sqrt(2))i,a=-1/(2sqrt(2))i

Find the zeros of 8a^2-4=-5

The zeros of 8a^2-4=-5 are a= 1/(2sqrt(2))i,a=-1/(2sqrt(2))i

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8 a^{2} - 4 = - 5

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

Solution steps

8 a^{2} - 4 = - 5

Move

4

to the right side

8 a^{2} = - 1

Divide both sides by

8

a^{2} = - \frac{1}{8}

For

x^{2} = f (a)

the solutions are

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

a = - \frac{1}{8}, a = - - \frac{1}{8}

- \frac{1}{8} = \frac{1}{2 2} i

- - \frac{1}{8} = - \frac{1}{2 2} i

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

\frac{( x - 4 )}{5 x + 7} \leq 2

125 q^{3} - w^{3}

a^{2} - 2 a - 8

27 x^{3} - 1

8 \cdot 7

What are the solutions to the equation 8a^2-4=-5 ?
The solutions to the equation 8a^2-4=-5 are a= 1/(2sqrt(2))i,a=-1/(2sqrt(2))i
Find the zeros of 8a^2-4=-5
The zeros of 8a^2-4=-5 are a= 1/(2sqrt(2))i,a=-1/(2sqrt(2))i