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

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

Find the zeros of z^2+z=1

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

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z^{2} + z = 1

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

Decimal Notation

z = 0.61803 \dots, z = - 1.61803 \dots

Solution steps

z^{2} + z = 1

Move

1

to the left side

z^{2} + z - 1 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

4 (x + 12)^{2} = 28

x, 6 x^{2} + 2 y^{2} - 4 = 2 x^{2} y + 3 y

4 x^{2} + 7 x + 10 = 0

x, z = x^{2} + y^{2} - 1

- 3 x^{2} = 4 x - 1

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