What are the solutions to the equation 6x^2-4x+3=0 ?

The solutions to the equation 6x^2-4x+3=0 are x= 1/3+i(sqrt(14))/6 ,x= 1/3-i(sqrt(14))/6

Find the zeros of 6x^2-4x+3=0

The zeros of 6x^2-4x+3=0 are x= 1/3+i(sqrt(14))/6 ,x= 1/3-i(sqrt(14))/6

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6 x^{2} - 4 x + 3 = 0

x = \frac{1}{3} + i \frac{14}{6}, x = \frac{1}{3} - i \frac{14}{6}

Solution steps

6 x^{2} - 4 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 6 \cdot 3}{2 \cdot 6}

Simplify

(- 4)^{2} - 4 \cdot 6 \cdot 3 : 214 i

x_{1, 2} = \frac{- ( - 4 ) \pm 2 14 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 14 i}{2 \cdot 6}, x_{2} = \frac{- ( - 4 ) - 2 14 i}{2 \cdot 6}

x = \frac{- ( - 4 ) + 2 14 i}{2 \cdot 6} : \frac{1}{3} + i \frac{14}{6}

x = \frac{- ( - 4 ) - 2 14 i}{2 \cdot 6} : \frac{1}{3} - i \frac{14}{6}

The solutions to the quadratic equation are:

x = \frac{1}{3} + i \frac{14}{6}, x = \frac{1}{3} - i \frac{14}{6}

x^{2} - 2 x + 17 = 0

t, s = 5 + 3 t + 2 t^{2}

y^{2} - 63 = 0

- \frac{3}{4} x^{2} + 9 x - 34 = 0

5 (x + 2)^{2} + 125 = 0

What are the solutions to the equation 6x^2-4x+3=0 ?
The solutions to the equation 6x^2-4x+3=0 are x= 1/3+i(sqrt(14))/6 ,x= 1/3-i(sqrt(14))/6
Find the zeros of 6x^2-4x+3=0
The zeros of 6x^2-4x+3=0 are x= 1/3+i(sqrt(14))/6 ,x= 1/3-i(sqrt(14))/6