What are the solutions to the equation 4+9n-3n^2=2-n ?

The solutions to the equation 4+9n-3n^2=2-n are n=-(-5+sqrt(31))/3 ,n=(5+sqrt(31))/3

Find the zeros of 4+9n-3n^2=2-n

The zeros of 4+9n-3n^2=2-n are n=-(-5+sqrt(31))/3 ,n=(5+sqrt(31))/3

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4 + 9 n - 3 n^{2} = 2 - n

n = - \frac{- 5 + 31}{3}, n = \frac{5 + 31}{3}

Decimal Notation

n = - 0.18925 \dots, n = 3.52258 \dots

Solution steps

4 + 9 n - 3 n^{2} = 2 - n

Move

n

to the left side

- 3 n^{2} + 10 n + 4 = 2

Move

2

to the left side

- 3 n^{2} + 10 n + 2 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

1 0^{2} - 4 (- 3) \cdot 2 = 231

n_{1, 2} = \frac{- 10 \pm 2 31}{2 ( - 3 )}

Separate the solutions

n_{1} = \frac{- 10 + 2 31}{2 ( - 3 )}, n_{2} = \frac{- 10 - 2 31}{2 ( - 3 )}

n = \frac{- 10 + 2 31}{2 ( - 3 )} : - \frac{- 5 + 31}{3}

n = \frac{- 10 - 2 31}{2 ( - 3 )} : \frac{5 + 31}{3}

The solutions to the quadratic equation are:

n = - \frac{- 5 + 31}{3}, n = \frac{5 + 31}{3}

25 x^{2} = 0

0 = 12 x^{2} - 32 x

x^{2} - 4 x - 3 = 2

9 x^{2} + 6 x + 1

x + 15 = x (x - 1)

What are the solutions to the equation 4+9n-3n^2=2-n ?
The solutions to the equation 4+9n-3n^2=2-n are n=-(-5+sqrt(31))/3 ,n=(5+sqrt(31))/3
Find the zeros of 4+9n-3n^2=2-n
The zeros of 4+9n-3n^2=2-n are n=-(-5+sqrt(31))/3 ,n=(5+sqrt(31))/3