What are the solutions to the equation n^2+n/3 =-180 ?

The solutions to the equation n^2+n/3 =-180 are n=-1/6+i(sqrt(6479))/6 ,n=-1/6-i(sqrt(6479))/6

Find the zeros of n^2+n/3 =-180

The zeros of n^2+n/3 =-180 are n=-1/6+i(sqrt(6479))/6 ,n=-1/6-i(sqrt(6479))/6

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n^{2} + \frac{n}{3} = - 180

n = - \frac{1}{6} + i \frac{6479}{6}, n = - \frac{1}{6} - i \frac{6479}{6}

Solution steps

n^{2} + \frac{n}{3} = - 180

Multiply both sides by

3

3 n^{2} + n = - 540

Move

540

to the left side

3 n^{2} + n + 540 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 \cdot 540}{2 \cdot 3}

Simplify

1^{2} - 4 \cdot 3 \cdot 540 : 6479 i

n_{1, 2} = \frac{- 1 \pm 6479 i}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 1 + 6479 i}{2 \cdot 3}, n_{2} = \frac{- 1 - 6479 i}{2 \cdot 3}

n = \frac{- 1 + 6479 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{6479}{6}

n = \frac{- 1 - 6479 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{6479}{6}

The solutions to the quadratic equation are:

n = - \frac{1}{6} + i \frac{6479}{6}, n = - \frac{1}{6} - i \frac{6479}{6}

2 x^{2} - 6 x + 2

6 x^{2} - x - 2 = 0

7 y^{2} + 23 y - 20 = 0

(x - 1) \cdot x = 12

x^{2} - 60 x + 500 = 0

What are the solutions to the equation n^2+n/3 =-180 ?
The solutions to the equation n^2+n/3 =-180 are n=-1/6+i(sqrt(6479))/6 ,n=-1/6-i(sqrt(6479))/6
Find the zeros of n^2+n/3 =-180
The zeros of n^2+n/3 =-180 are n=-1/6+i(sqrt(6479))/6 ,n=-1/6-i(sqrt(6479))/6