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

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

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

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

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3 x^{2} + x = - 2

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

Solution steps

3 x^{2} + x = - 2

Move

2

to the left side

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}

1 + x + x^{2} = a (1 + x) + b \cdot (1 + x^{2}) + c \cdot (x + x^{2})

x, 16 x^{2} - 24 x y + 9 y^{2} - 30 x - 40 y = 0

2 x^{2} - 3 x + 20 = 0

x^{2} - 3 x - 7 = 3

- 9 y^{2} + 24 = 0

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