What are the solutions to the equation 12x^2+48=0 ?

The solutions to the equation 12x^2+48=0 are x=2i,x=-2i

Find the zeros of 12x^2+48=0

The zeros of 12x^2+48=0 are x=2i,x=-2i

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12 x^{2} + 48 = 0

x = 2 i, x = - 2 i

Solution steps

12 x^{2} + 48 = 0

Move

48

to the right side

12 x^{2} = - 48

Divide both sides by

12

x^{2} = - 4

For

x^{2} = f (a)

the solutions are

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

x = - 4, x = - - 4

- 4 = 2 i

- - 4 = - 2 i

x = 2 i, x = - 2 i

h^{2} - 3 h - 28 = 0

(x + 5)^{2} + (x + 3)^{2} = (2 x + 4)^{2}

4 x^{2} + 19 x + 12 = - 5

x^{2} - 1 = 2 x + 7

0 = 6 x^{2} - 8 x

What are the solutions to the equation 12x^2+48=0 ?
The solutions to the equation 12x^2+48=0 are x=2i,x=-2i
Find the zeros of 12x^2+48=0
The zeros of 12x^2+48=0 are x=2i,x=-2i