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

The solutions to the equation 3x^2+3x=2x-1 are x=-1/6+i(sqrt(11))/6 ,x=-1/6-i(sqrt(11))/6

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

The zeros of 3x^2+3x=2x-1 are x=-1/6+i(sqrt(11))/6 ,x=-1/6-i(sqrt(11))/6

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

x = - \frac{1}{6} + i \frac{11}{6}, x = - \frac{1}{6} - i \frac{11}{6}

Solution steps

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

Move

1

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 3 \cdot 1 : 11 i

x_{1, 2} = \frac{- 1 \pm 11 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 1 + 11 i}{2 \cdot 3}, x_{2} = \frac{- 1 - 11 i}{2 \cdot 3}

x = \frac{- 1 + 11 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{11}{6}

x = \frac{- 1 - 11 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{11}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{11}{6}, x = - \frac{1}{6} - i \frac{11}{6}

(λ^{2} - 14) - 20 = λ

x, x^{2} - 2 x = m^{2} + 2 m

- 3 x^{2} + (7 k^{2} - 3) x + 7 = 0

w, z = - 2 w^{2}

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

What are the solutions to the equation 3x^2+3x=2x-1 ?
The solutions to the equation 3x^2+3x=2x-1 are x=-1/6+i(sqrt(11))/6 ,x=-1/6-i(sqrt(11))/6
Find the zeros of 3x^2+3x=2x-1
The zeros of 3x^2+3x=2x-1 are x=-1/6+i(sqrt(11))/6 ,x=-1/6-i(sqrt(11))/6