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 = 1.61803 \dots, x = - 0.61803 \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}

(- 4 x)^{2}

a^{2} b^{2} + ab

4 a^{2} - 2

\frac{4 x y}{x - 2} \cdot \frac{x ^{2}}{y ^{2} x}

e^{1^{2}}

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