What are the solutions to the equation x^2-8+12=0 ?

The solutions to the equation x^2-8+12=0 are x=2i,x=-2i

Find the zeros of x^2-8+12=0

The zeros of x^2-8+12=0 are x=2i,x=-2i

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x^{2} - 8 + 12 = 0

x = 2 i, x = - 2 i

Solution steps

x^{2} - 8 + 12 = 0

Move

8

to the right side

x^{2} + 12 = 8

Move

12

to the right side

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

\frac{30}{100} \cdot 50

- 7 y < 35

\frac{x}{4} + x = \frac{3}{8}

(m^{4})^{\frac{3}{2}}

\frac{3 x}{10} + 1 \geq \frac{1}{5} - \frac{x}{10}

What are the solutions to the equation x^2-8+12=0 ?
The solutions to the equation x^2-8+12=0 are x=2i,x=-2i
Find the zeros of x^2-8+12=0
The zeros of x^2-8+12=0 are x=2i,x=-2i