What are the solutions to the equation n^2-n=-41 ?

The solutions to the equation n^2-n=-41 are n= 1/2+i(sqrt(163))/2 ,n= 1/2-i(sqrt(163))/2

Find the zeros of n^2-n=-41

The zeros of n^2-n=-41 are n= 1/2+i(sqrt(163))/2 ,n= 1/2-i(sqrt(163))/2

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n^{2} - n = - 41

n = \frac{1}{2} + i \frac{163}{2}, n = \frac{1}{2} - i \frac{163}{2}

Solution steps

n^{2} - n = - 41

Move

41

to the left side

n^{2} - n + 41 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 41 : 163 i

n_{1, 2} = \frac{- ( - 1 ) \pm 163 i}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 163 i}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 163 i}{2 \cdot 1}

n = \frac{- ( - 1 ) + 163 i}{2 \cdot 1} : \frac{1}{2} + i \frac{163}{2}

n = \frac{- ( - 1 ) - 163 i}{2 \cdot 1} : \frac{1}{2} - i \frac{163}{2}

The solutions to the quadratic equation are:

n = \frac{1}{2} + i \frac{163}{2}, n = \frac{1}{2} - i \frac{163}{2}

y, x^{3} y^{2} - 5 x^{2} y + x = 1

x^{2} - 25 x - 5 = 0

2 x^{2} - 16 x - 7 = 0

3 y^{2} - 10 y = - 5

3 x^{2} - 4 a x - 2 a^{2} = 0

What are the solutions to the equation n^2-n=-41 ?
The solutions to the equation n^2-n=-41 are n= 1/2+i(sqrt(163))/2 ,n= 1/2-i(sqrt(163))/2
Find the zeros of n^2-n=-41
The zeros of n^2-n=-41 are n= 1/2+i(sqrt(163))/2 ,n= 1/2-i(sqrt(163))/2