What are the solutions to the equation 3x^2+14=12 ?

The solutions to the equation 3x^2+14=12 are x=sqrt(2/3)i,x=-sqrt(2/3)i

Find the zeros of 3x^2+14=12

The zeros of 3x^2+14=12 are x=sqrt(2/3)i,x=-sqrt(2/3)i

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3 x^{2} + 14 = 12

x = \frac{2}{3} i, x = - \frac{2}{3} i

Solution steps

3 x^{2} + 14 = 12

Move

14

to the right side

3 x^{2} = - 2

Divide both sides by

3

x^{2} = - \frac{2}{3}

For

x^{2} = f (a)

the solutions are

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

x = - \frac{2}{3}, x = - - \frac{2}{3}

- \frac{2}{3} = \frac{2}{3} i

- - \frac{2}{3} = - \frac{2}{3} i

x = \frac{2}{3} i, x = - \frac{2}{3} i

\frac{3}{7} + \frac{13}{4} \cdot \frac{1}{3}

lo g_{25} (49)

\frac{( 2 x - 3 )}{6} - 2 (\frac{3}{2} x - 1) - \frac{3 - 2 x}{3} = 0

- 45 x^{8} y^{2} z^{4} - 18 x^{4} y^{6} z^{7}

3 x + 1 = 8 x - 24

What are the solutions to the equation 3x^2+14=12 ?
The solutions to the equation 3x^2+14=12 are x=sqrt(2/3)i,x=-sqrt(2/3)i
Find the zeros of 3x^2+14=12
The zeros of 3x^2+14=12 are x=sqrt(2/3)i,x=-sqrt(2/3)i