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

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

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

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

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x^{2} - x + 1 = 2

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

Decimal Notation

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

Solution steps

x^{2} - x + 1 = 2

Move

2

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 x^{2} - 4 x + 2 = 0

1 + 0.2 x + 0.7 x^{2} = 0

\frac{1}{6} n^{2} - \frac{1}{2} n - \frac{2}{3} = 0

(x + 3)^{2} + 3 x^{2} = 10 x + 8

x^{2} + 70 x - 3000 = 0

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