What are the solutions to the equation x^2-5x+1=0 ?

The solutions to the equation x^2-5x+1=0 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

Find the zeros of x^2-5x+1=0

The zeros of x^2-5x+1=0 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2

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x^{2} - 5 x + 1 = 0

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

Decimal Notation

x = 4.79128 \dots, x = 0.20871 \dots

Solution steps

x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 1 = 21

x_{1, 2} = \frac{- ( - 5 ) \pm 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 21}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 21}{2 \cdot 1}

x = \frac{- ( - 5 ) + 21}{2 \cdot 1} : \frac{5 + 21}{2}

x = \frac{- ( - 5 ) - 21}{2 \cdot 1} : \frac{5 - 21}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 21}{2}, x = \frac{5 - 21}{2}

(4 - 3 i) + (2 i - 8)

- 2 (x + 1) < 2

32 t^{2} - 18 u^{2}

- 2 (4 y - 7) + y \geq 2 y - (- 5 + y)

56 \cdot 9 \div 14 \cdot \frac{1}{3}

What are the solutions to the equation x^2-5x+1=0 ?
The solutions to the equation x^2-5x+1=0 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2
Find the zeros of x^2-5x+1=0
The zeros of x^2-5x+1=0 are x=(5+sqrt(21))/2 ,x=(5-sqrt(21))/2