What are the solutions to the equation 4x^2+12=-8 ?

The solutions to the equation 4x^2+12=-8 are x=sqrt(5)i,x=-sqrt(5)i

Find the zeros of 4x^2+12=-8

The zeros of 4x^2+12=-8 are x=sqrt(5)i,x=-sqrt(5)i

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4 x^{2} + 12 = - 8

x = 5 i, x = - 5 i

Solution steps

4 x^{2} + 12 = - 8

Move

12

to the right side

4 x^{2} = - 20

Divide both sides by

4

x^{2} = - 5

For

x^{2} = f (a)

the solutions are

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

x = - 5, x = - - 5

- 5 = 5 i

- - 5 = - 5 i

x = 5 i, x = - 5 i

x^{2} - 16 x - 612 = 0

x^{2} + 6 x - 4 = x + 2

p^{2} + 4 p + 6 = 0

x, x^{2} + \frac{k}{m} x = 0

r, x^{2} r^{2} - 3 x r + 3 = 0

What are the solutions to the equation 4x^2+12=-8 ?
The solutions to the equation 4x^2+12=-8 are x=sqrt(5)i,x=-sqrt(5)i
Find the zeros of 4x^2+12=-8
The zeros of 4x^2+12=-8 are x=sqrt(5)i,x=-sqrt(5)i