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

The solutions to the equation 3x^2-x+6=0 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6

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

The zeros of 3x^2-x+6=0 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6

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

x = \frac{1}{6} + i \frac{71}{6}, x = \frac{1}{6} - i \frac{71}{6}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 3 \cdot 6 : 71 i

x_{1, 2} = \frac{- ( - 1 ) \pm 71 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 71 i}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 71 i}{2 \cdot 3}

x = \frac{- ( - 1 ) + 71 i}{2 \cdot 3} : \frac{1}{6} + i \frac{71}{6}

x = \frac{- ( - 1 ) - 71 i}{2 \cdot 3} : \frac{1}{6} - i \frac{71}{6}

The solutions to the quadratic equation are:

x = \frac{1}{6} + i \frac{71}{6}, x = \frac{1}{6} - i \frac{71}{6}

5 (2 - x) - x = 2 (x + 1)

\frac{x + 1}{3} + \frac{x}{5} = \frac{x + 1}{2}

\frac{( 9 b ^{2} + 33 b - 18 )}{( b + 4 )}

x^{2} \leq 4 x + 12

What are the solutions to the equation 3x^2-x+6=0 ?
The solutions to the equation 3x^2-x+6=0 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6
Find the zeros of 3x^2-x+6=0
The zeros of 3x^2-x+6=0 are x= 1/6+i(sqrt(71))/6 ,x= 1/6-i(sqrt(71))/6