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

The solutions to the equation 1/2 n^2+2/3 n=-1 are n=-2/3+i(sqrt(14))/3 ,n=-2/3-i(sqrt(14))/3

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

The zeros of 1/2 n^2+2/3 n=-1 are n=-2/3+i(sqrt(14))/3 ,n=-2/3-i(sqrt(14))/3

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

n = - \frac{2}{3} + i \frac{14}{3}, n = - \frac{2}{3} - i \frac{14}{3}

Solution steps

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

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

\frac{1}{2} n^{2} \cdot 6 + \frac{2}{3} n \cdot 6 = - 1 \cdot 6

Simplify

3 n^{2} + 4 n = - 6

Move

6

to the left side

3 n^{2} + 4 n + 6 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

Simplify

4^{2} - 4 \cdot 3 \cdot 6 : 214 i

n_{1, 2} = \frac{- 4 \pm 2 14 i}{2 \cdot 3}

Separate the solutions

n_{1} = \frac{- 4 + 2 14 i}{2 \cdot 3}, n_{2} = \frac{- 4 - 2 14 i}{2 \cdot 3}

n = \frac{- 4 + 2 14 i}{2 \cdot 3} : - \frac{2}{3} + i \frac{14}{3}

n = \frac{- 4 - 2 14 i}{2 \cdot 3} : - \frac{2}{3} - i \frac{14}{3}

The solutions to the quadratic equation are:

n = - \frac{2}{3} + i \frac{14}{3}, n = - \frac{2}{3} - i \frac{14}{3}

2 x^{2} + 8 x + 7

(0.25 - x)^{2} = 2

4 x^{2} + 8 x - 16 = 0

4 x^{2} + 2 = 9 x

2 v^{2} + 5 v - 7 = 0

What are the solutions to the equation 1/2 n^2+2/3 n=-1 ?
The solutions to the equation 1/2 n^2+2/3 n=-1 are n=-2/3+i(sqrt(14))/3 ,n=-2/3-i(sqrt(14))/3
Find the zeros of 1/2 n^2+2/3 n=-1
The zeros of 1/2 n^2+2/3 n=-1 are n=-2/3+i(sqrt(14))/3 ,n=-2/3-i(sqrt(14))/3