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

The solutions to the equation 3x^2+x+2=0 are x=-1/6+i(sqrt(23))/6 ,x=-1/6-i(sqrt(23))/6

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

The zeros of 3x^2+x+2=0 are x=-1/6+i(sqrt(23))/6 ,x=-1/6-i(sqrt(23))/6

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

x = - \frac{1}{6} + i \frac{23}{6}, x = - \frac{1}{6} - i \frac{23}{6}

Solution steps

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 3 \cdot 2 : 23 i

x_{1, 2} = \frac{- 1 \pm 23 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 23 i}{2 \cdot 3}, x_{2} = \frac{- 1 - 23 i}{2 \cdot 3}

x = \frac{- 1 + 23 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{23}{6}

x = \frac{- 1 - 23 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{23}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{23}{6}, x = - \frac{1}{6} - i \frac{23}{6}

5 x^{2} - 20 x + 20

\frac{1}{8} (2 x + 4) > \frac{1}{4} (2 x + 8) + 1

48 n^{3} + 9 n - 3 n

\frac{9}{6} = \frac{x}{10}

- 16 t^{2} + 64 t + 80 = 0

What are the solutions to the equation 3x^2+x+2=0 ?
The solutions to the equation 3x^2+x+2=0 are x=-1/6+i(sqrt(23))/6 ,x=-1/6-i(sqrt(23))/6
Find the zeros of 3x^2+x+2=0
The zeros of 3x^2+x+2=0 are x=-1/6+i(sqrt(23))/6 ,x=-1/6-i(sqrt(23))/6