What are the solutions to the equation -5n^2+2n-7=0 ?

The solutions to the equation -5n^2+2n-7=0 are n= 1/5-i(sqrt(34))/5 ,n= 1/5+i(sqrt(34))/5

Find the zeros of -5n^2+2n-7=0

The zeros of -5n^2+2n-7=0 are n= 1/5-i(sqrt(34))/5 ,n= 1/5+i(sqrt(34))/5

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

n = \frac{1}{5} - i \frac{34}{5}, n = \frac{1}{5} + i \frac{34}{5}

Solution steps

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

Solve with the quadratic formula

n_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) ( - 7 )}{2 ( - 5 )}

Simplify

2^{2} - 4 (- 5) (- 7) : 234 i

n_{1, 2} = \frac{- 2 \pm 2 34 i}{2 ( - 5 )}

Separate the solutions

n_{1} = \frac{- 2 + 2 34 i}{2 ( - 5 )}, n_{2} = \frac{- 2 - 2 34 i}{2 ( - 5 )}

n = \frac{- 2 + 2 34 i}{2 ( - 5 )} : \frac{1}{5} - i \frac{34}{5}

n = \frac{- 2 - 2 34 i}{2 ( - 5 )} : \frac{1}{5} + i \frac{34}{5}

The solutions to the quadratic equation are:

n = \frac{1}{5} - i \frac{34}{5}, n = \frac{1}{5} + i \frac{34}{5}

2 x^{2} + 7 x = 3

4 + 16.3 t - 4.9 t^{2} = 0

(x - 20)^{2} = 81

4 t^{2} - t - 5 = 0

z^{2} = 38

What are the solutions to the equation -5n^2+2n-7=0 ?
The solutions to the equation -5n^2+2n-7=0 are n= 1/5-i(sqrt(34))/5 ,n= 1/5+i(sqrt(34))/5
Find the zeros of -5n^2+2n-7=0
The zeros of -5n^2+2n-7=0 are n= 1/5-i(sqrt(34))/5 ,n= 1/5+i(sqrt(34))/5