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

The solutions to the equation 3n^2+2n-1=-2n are n=(-2+sqrt(7))/3 ,n=-(2+sqrt(7))/3

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

The zeros of 3n^2+2n-1=-2n are n=(-2+sqrt(7))/3 ,n=-(2+sqrt(7))/3

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

n = \frac{- 2 + 7}{3}, n = - \frac{2 + 7}{3}

Decimal Notation

n = 0.21525 \dots, n = - 1.54858 \dots

Solution steps

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

Move

2 n

to the left side

3 n^{2} + 4 n - 1 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

4^{2} - 4 \cdot 3 (- 1) = 27

n_{1, 2} = \frac{- 4 \pm 2 7}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 4 + 2 7}{2 \cdot 3}, n_{2} = \frac{- 4 - 2 7}{2 \cdot 3}

n = \frac{- 4 + 2 7}{2 \cdot 3} : \frac{- 2 + 7}{3}

n = \frac{- 4 - 2 7}{2 \cdot 3} : - \frac{2 + 7}{3}

The solutions to the quadratic equation are:

n = \frac{- 2 + 7}{3}, n = - \frac{2 + 7}{3}

3 k^{2} + 2 = 0

0 = 3 x^{2} + 6

x^{2} - 7 x + 15 = 0

x^{2} - 14 x + 118 = 5

4 x^{2} + 12 x + 13 = 0

What are the solutions to the equation 3n^2+2n-1=-2n ?
The solutions to the equation 3n^2+2n-1=-2n are n=(-2+sqrt(7))/3 ,n=-(2+sqrt(7))/3
Find the zeros of 3n^2+2n-1=-2n
The zeros of 3n^2+2n-1=-2n are n=(-2+sqrt(7))/3 ,n=-(2+sqrt(7))/3