What are the solutions to the equation solvefor x,x^2+x-1=0 ?

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

Find the zeros of solvefor x,x^2+x-1=0

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

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x, x^{2} + x - 1 = 0

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

Decimal Notation

x = 0.61803 \dots, x = - 1.61803 \dots

Solution steps

x^{2} + x - 1 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 5}{2 \cdot 1}, x_{2} = \frac{- 1 - 5}{2 \cdot 1}

x = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

x = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

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

1.77 \cdot 1 0^{- 6} - 1.77 \cdot 1 0^{- 5} x - x^{2} = 0

\frac{1}{5} (x + 3)^{2} - 2 = 18

32 x^{2} - 16 x = 0

4 x^{2} - x = 3

x^{2} + x^{2} = 2^{4}

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