What are the solutions to the equation 3x^2+2x=-1 ?

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

Find the zeros of 3x^2+2x=-1

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

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3 x^{2} + 2 x = - 1

x = - \frac{1}{3} + i \frac{2}{3}, x = - \frac{1}{3} - i \frac{2}{3}

Solution steps

3 x^{2} + 2 x = - 1

Move

1

to the left side

3 x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 3 \cdot 1 : 22 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - \frac{1}{3} + i \frac{2}{3}, x = - \frac{1}{3} - i \frac{2}{3}

(4 x + 1)^{2} - 8 = 1

x, x^{2} - 4 x + 5 = 0

v, \frac{1}{2} m v^{2} = m g h

x^{2} + 0.13 x + 0.0032 = 0

y, 2 x^{3} + x^{2} y + y^{2} = 4

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