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

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

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

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

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0 = x^{2} + x - 1

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

Decimal Notation

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

Solution steps

0 = x^{2} + x - 1

Switch sides

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}

- 7 x < 42

(x + 2)^{2} = 4

2 a^{2} + 11 a + 5

\frac{2 ^{0}}{( 2 ^{- 2} x ^{\frac{1}{2}} y ^{- 2} ) ^{3}}

\frac{\frac{2}{3}}{\frac{1}{5}}

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