What are the solutions to the equation n^2-n+1=17999 ?

The solutions to the equation n^2-n+1=17999 are n=(1+sqrt(71993))/2 ,n=(1-sqrt(71993))/2

Find the zeros of n^2-n+1=17999

The zeros of n^2-n+1=17999 are n=(1+sqrt(71993))/2 ,n=(1-sqrt(71993))/2

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n^{2} - n + 1 = 17999

n = \frac{1 + 71993}{2}, n = \frac{1 - 71993}{2}

Decimal Notation

n = 134.65755 \dots, n = - 133.65755 \dots

Solution steps

n^{2} - n + 1 = 17999

Move

17999

to the left side

n^{2} - n - 17998 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 17998 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 17998) = 71993

n_{1, 2} = \frac{- ( - 1 ) \pm 71993}{2 \cdot 1}

Separate the solutions

n_{1} = \frac{- ( - 1 ) + 71993}{2 \cdot 1}, n_{2} = \frac{- ( - 1 ) - 71993}{2 \cdot 1}

n = \frac{- ( - 1 ) + 71993}{2 \cdot 1} : \frac{1 + 71993}{2}

n = \frac{- ( - 1 ) - 71993}{2 \cdot 1} : \frac{1 - 71993}{2}

The solutions to the quadratic equation are:

n = \frac{1 + 71993}{2}, n = \frac{1 - 71993}{2}

8 x^{2} + 189 x - 72 = 0

9 x^{2} + 7 x + 49 = 0

13 x - 3 x^{2} = 0

x^{2} = 242

15 x - 10 = 6 x - (x x^{2}) + (- x + 3)

What are the solutions to the equation n^2-n+1=17999 ?
The solutions to the equation n^2-n+1=17999 are n=(1+sqrt(71993))/2 ,n=(1-sqrt(71993))/2
Find the zeros of n^2-n+1=17999
The zeros of n^2-n+1=17999 are n=(1+sqrt(71993))/2 ,n=(1-sqrt(71993))/2