What are the solutions to the equation (-n)=3(-n)^2-(-n) ?

The solutions to the equation (-n)=3(-n)^2-(-n) are n=0,n=-2/3

Find the zeros of (-n)=3(-n)^2-(-n)

The zeros of (-n)=3(-n)^2-(-n) are n=0,n=-2/3

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(- n) = 3 (- n)^{2} - (- n)

n = 0, n = - \frac{2}{3}

Decimal Notation

n = 0, n = - 0.66666 \dots

Solution steps

(- n) = 3 (- n)^{2} - (- n)

Expand

3 (- n)^{2} - (- n) : 3 n^{2} + n

- n = 3 n^{2} + n

Switch sides

3 n^{2} + n = - n

Move

n

to the left side

3 n^{2} + 2 n = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

2^{2} - 4 \cdot 3 \cdot 0 = 2

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

Separate the solutions

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

n = \frac{- 2 + 2}{2 \cdot 3} : 0

n = \frac{- 2 - 2}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

n = 0, n = - \frac{2}{3}

x^{2} - 1 - 3 = 0

x^{2} = 27 - 6 x

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

2 m^{2} - 10 = 36

5 (14 + 5) = s^{2} - 7^{2}

What are the solutions to the equation (-n)=3(-n)^2-(-n) ?
The solutions to the equation (-n)=3(-n)^2-(-n) are n=0,n=-2/3
Find the zeros of (-n)=3(-n)^2-(-n)
The zeros of (-n)=3(-n)^2-(-n) are n=0,n=-2/3