What are the solutions to the equation 6x^2-3x+7=0 ?

The solutions to the equation 6x^2-3x+7=0 are x= 1/4+i(sqrt(159))/(12),x= 1/4-i(sqrt(159))/(12)

Find the zeros of 6x^2-3x+7=0

The zeros of 6x^2-3x+7=0 are x= 1/4+i(sqrt(159))/(12),x= 1/4-i(sqrt(159))/(12)

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6 x^{2} - 3 x + 7 = 0

x = \frac{1}{4} + i \frac{159}{12}, x = \frac{1}{4} - i \frac{159}{12}

Solution steps

6 x^{2} - 3 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

Simplify

(- 3)^{2} - 4 \cdot 6 \cdot 7 : 159 i

x_{1, 2} = \frac{- ( - 3 ) \pm 159 i}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 159 i}{2 \cdot 6}, x_{2} = \frac{- ( - 3 ) - 159 i}{2 \cdot 6}

x = \frac{- ( - 3 ) + 159 i}{2 \cdot 6} : \frac{1}{4} + i \frac{159}{12}

x = \frac{- ( - 3 ) - 159 i}{2 \cdot 6} : \frac{1}{4} - i \frac{159}{12}

The solutions to the quadratic equation are:

x = \frac{1}{4} + i \frac{159}{12}, x = \frac{1}{4} - i \frac{159}{12}

x, x^{2} + 9 y^{2} = k

\frac{x ^{2}}{4} = 16

x^{2} - 4 x - 50 = 0

17 x^{2} - 6 x = 0

- 6 x^{2} + 20 x = 0

What are the solutions to the equation 6x^2-3x+7=0 ?
The solutions to the equation 6x^2-3x+7=0 are x= 1/4+i(sqrt(159))/(12),x= 1/4-i(sqrt(159))/(12)
Find the zeros of 6x^2-3x+7=0
The zeros of 6x^2-3x+7=0 are x= 1/4+i(sqrt(159))/(12),x= 1/4-i(sqrt(159))/(12)