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}

11 = t + 4

\frac{3}{2 + x} < \frac{1}{x - 1}

\frac{5 n ^{8}}{20 n ^{8}}

\frac{1}{2} + \frac{2}{3} - \frac{7}{4}

\frac{3}{5} x - 7 \leq 5 x - \frac{9}{5}

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