What are the solutions to the equation 12x^2-4x+1=0 ?

The solutions to the equation 12x^2-4x+1=0 are x= 1/6+i(sqrt(2))/6 ,x= 1/6-i(sqrt(2))/6

Find the zeros of 12x^2-4x+1=0

The zeros of 12x^2-4x+1=0 are x= 1/6+i(sqrt(2))/6 ,x= 1/6-i(sqrt(2))/6

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12 x^{2} - 4 x + 1 = 0

x = \frac{1}{6} + i \frac{2}{6}, x = \frac{1}{6} - i \frac{2}{6}

Solution steps

12 x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

Simplify

(- 4)^{2} - 4 \cdot 12 \cdot 1 : 42 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 2 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 2 i}{2 \cdot 12}, x_{2} = \frac{- ( - 4 ) - 4 2 i}{2 \cdot 12}

x = \frac{- ( - 4 ) + 4 2 i}{2 \cdot 12} : \frac{1}{6} + i \frac{2}{6}

x = \frac{- ( - 4 ) - 4 2 i}{2 \cdot 12} : \frac{1}{6} - i \frac{2}{6}

The solutions to the quadratic equation are:

x = \frac{1}{6} + i \frac{2}{6}, x = \frac{1}{6} - i \frac{2}{6}

x^{2} + 4 x - 28 = 0

12 x^{2} + 5 x = 3

x + 2 x^{2} = 0

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

4 (3 x - 8)^{2} = 64

What are the solutions to the equation 12x^2-4x+1=0 ?
The solutions to the equation 12x^2-4x+1=0 are x= 1/6+i(sqrt(2))/6 ,x= 1/6-i(sqrt(2))/6
Find the zeros of 12x^2-4x+1=0
The zeros of 12x^2-4x+1=0 are x= 1/6+i(sqrt(2))/6 ,x= 1/6-i(sqrt(2))/6