What are the solutions to the equation 12x^2-4=0 ?

The solutions to the equation 12x^2-4=0 are x= 1/(sqrt(3)),x=-1/(sqrt(3))

Find the zeros of 12x^2-4=0

The zeros of 12x^2-4=0 are x= 1/(sqrt(3)),x=-1/(sqrt(3))

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12 x^{2} - 4 = 0

x = \frac{1}{3}, x = - \frac{1}{3}

Decimal Notation

x = 0.57735 \dots, x = - 0.57735 \dots

Solution steps

12 x^{2} - 4 = 0

Move

4

to the right side

12 x^{2} = 4

Divide both sides by

12

x^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

x = \frac{1}{3}, x = - \frac{1}{3}

\frac{1}{3} = \frac{1}{3}

- \frac{1}{3} = - \frac{1}{3}

x = \frac{1}{3}, x = - \frac{1}{3}

\frac{2 r + 5 + r + 4}{2} + 3 > 6

- 2 (3)^{2}

(c^{4} - c^{2} + 1) (c^{4} + c^{2} + 1) = c^{8} + c^{4} + 1

18 h^{2} + 15 h - 18

\frac{1 + i}{i}

What are the solutions to the equation 12x^2-4=0 ?
The solutions to the equation 12x^2-4=0 are x= 1/(sqrt(3)),x=-1/(sqrt(3))
Find the zeros of 12x^2-4=0
The zeros of 12x^2-4=0 are x= 1/(sqrt(3)),x=-1/(sqrt(3))