What are the solutions to the equation 3x^2+12=0 ?

The solutions to the equation 3x^2+12=0 are x=2i,x=-2i

Find the zeros of 3x^2+12=0

The zeros of 3x^2+12=0 are x=2i,x=-2i

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3 x^{2} + 12 = 0

x = 2 i, x = - 2 i

Solution steps

3 x^{2} + 12 = 0

Move

12

to the right side

3 x^{2} = - 12

Divide both sides by

3

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

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

a^{2} - 18 a + 68 = 3

\frac{56}{8}

\frac{5}{- 3 + 4 i}

\frac{2 2}{2}

What are the solutions to the equation 3x^2+12=0 ?
The solutions to the equation 3x^2+12=0 are x=2i,x=-2i
Find the zeros of 3x^2+12=0
The zeros of 3x^2+12=0 are x=2i,x=-2i