What are the solutions to the equation 125x^2+20=0 ?

The solutions to the equation 125x^2+20=0 are x= 2/5 i,x=-2/5 i

Find the zeros of 125x^2+20=0

The zeros of 125x^2+20=0 are x= 2/5 i,x=-2/5 i

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125 x^{2} + 20 = 0

x = \frac{2}{5} i, x = - \frac{2}{5} i

Solution steps

125 x^{2} + 20 = 0

Move

20

to the right side

125 x^{2} = - 20

Divide both sides by

125

x^{2} = - \frac{4}{25}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{4}{25}, x = - - \frac{4}{25}

- \frac{4}{25} = \frac{2}{5} i

- - \frac{4}{25} = - \frac{2}{5} i

x = \frac{2}{5} i, x = - \frac{2}{5} i

x, 9 x^{2} + 16 y^{2} = 0

l (q) = 5 q^{2} - 8 q + 12

t, d t + 0.5 t = d t (0.5 t)

(7 - h)^{2} + (3 - k)^{2} = r^{2}

1 = t^{2}

What are the solutions to the equation 125x^2+20=0 ?
The solutions to the equation 125x^2+20=0 are x= 2/5 i,x=-2/5 i
Find the zeros of 125x^2+20=0
The zeros of 125x^2+20=0 are x= 2/5 i,x=-2/5 i