What are the solutions to the equation 6x^2-4x+12=0 ?

The solutions to the equation 6x^2-4x+12=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3

Find the zeros of 6x^2-4x+12=0

The zeros of 6x^2-4x+12=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3

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

x = \frac{1}{3} + i \frac{17}{3}, x = \frac{1}{3} - i \frac{17}{3}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

Simplify

(- 4)^{2} - 4 \cdot 6 \cdot 12 : 417 i

x_{1, 2} = \frac{- ( - 4 ) \pm 4 17 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 17 i}{2 \cdot 6}, x_{2} = \frac{- ( - 4 ) - 4 17 i}{2 \cdot 6}

x = \frac{- ( - 4 ) + 4 17 i}{2 \cdot 6} : \frac{1}{3} + i \frac{17}{3}

x = \frac{- ( - 4 ) - 4 17 i}{2 \cdot 6} : \frac{1}{3} - i \frac{17}{3}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{17}{3}, x = \frac{1}{3} - i \frac{17}{3}

(x - 3)^{2} = 16 - 1 - x^{2} + 2 x

x^{2} = - 3 x - 4

x, x^{2} y^{1} + x y = 1

x, x^{2} + y^{2} + 2^{2} - 2 x + 4 y + 4 = 0

3 m^{2} - 8 m + 4 = 0

What are the solutions to the equation 6x^2-4x+12=0 ?
The solutions to the equation 6x^2-4x+12=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3
Find the zeros of 6x^2-4x+12=0
The zeros of 6x^2-4x+12=0 are x= 1/3+i(sqrt(17))/3 ,x= 1/3-i(sqrt(17))/3