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

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

Find the zeros of x^2+x=1

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

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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

x^{2} + x = 1

Move

1

to the left side

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}

- 10 - n < - 12 or 9 + 4 n \geq 21

∣ 2 x - 1 ∣ \leq 3 - x

2 x^{2} + x = 10

100 = 50 x - 5 x^{2}

12 - 7

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