What are the solutions to the equation n^2-n-5=2n^2 ?

The solutions to the equation n^2-n-5=2n^2 are n=-1/2-i(sqrt(19))/2 ,n=-1/2+i(sqrt(19))/2

Find the zeros of n^2-n-5=2n^2

The zeros of n^2-n-5=2n^2 are n=-1/2-i(sqrt(19))/2 ,n=-1/2+i(sqrt(19))/2

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n^{2} - n - 5 = 2 n^{2}

n = - \frac{1}{2} - i \frac{19}{2}, n = - \frac{1}{2} + i \frac{19}{2}

Solution steps

n^{2} - n - 5 = 2 n^{2}

Move

2 n^{2}

to the left side

- n^{2} - n - 5 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) ( - 5 )}{2 ( - 1 )}

Simplify

(- 1)^{2} - 4 (- 1) (- 5) : 19 i

n_{1, 2} = \frac{- ( - 1 ) \pm 19 i}{2 ( - 1 )}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 19 i}{2 ( - 1 )}, n_{2} = \frac{- ( - 1 ) - 19 i}{2 ( - 1 )}

n = \frac{- ( - 1 ) + 19 i}{2 ( - 1 )} : - \frac{1}{2} - i \frac{19}{2}

n = \frac{- ( - 1 ) - 19 i}{2 ( - 1 )} : - \frac{1}{2} + i \frac{19}{2}

The solutions to the quadratic equation are:

n = - \frac{1}{2} - i \frac{19}{2}, n = - \frac{1}{2} + i \frac{19}{2}

2 y^{2} = - 8

- 6 y^{2} + 5 y + 2 = 0

9 x^{2} - 3 x + 4 = 0

x^{2} - 1.2 x + 0.2 = 0

x, 5 x^{2} - 7 x = 1

What are the solutions to the equation n^2-n-5=2n^2 ?
The solutions to the equation n^2-n-5=2n^2 are n=-1/2-i(sqrt(19))/2 ,n=-1/2+i(sqrt(19))/2
Find the zeros of n^2-n-5=2n^2
The zeros of n^2-n-5=2n^2 are n=-1/2-i(sqrt(19))/2 ,n=-1/2+i(sqrt(19))/2