What are the solutions to the equation n^2-n-1=0 ?

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

Find the zeros of n^2-n-1=0

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

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n^{2} - n - 1 = 0

n = \frac{1 + 5}{2}, n = \frac{1 - 5}{2}

Decimal Notation

n = 1.61803 \dots, n = - 0.61803 \dots

Solution steps

n^{2} - n - 1 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Separate the solutions

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

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

n = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

The solutions to the quadratic equation are:

n = \frac{1 + 5}{2}, n = \frac{1 - 5}{2}

7 m^{2} + 2 m - 4 = 0

h, (7 - h)^{2} + (2 - k)^{2} = 0

x (x - 35) = 0

13 x^{2} - 39 = - 26 x

36 = x^{2} - 5 x

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