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(29))/2 ,x=(5-sqrt(29))/2

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

The zeros of x^2-5x-1=0 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2

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

x = \frac{5 + 29}{2}, x = \frac{5 - 29}{2}

Decimal Notation

x = 5.19258 \dots, x = - 0.19258 \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) = 29

x_{1, 2} = \frac{- ( - 5 ) \pm 29}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 29}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 29}{2 \cdot 1}

x = \frac{- ( - 5 ) + 29}{2 \cdot 1} : \frac{5 + 29}{2}

x = \frac{- ( - 5 ) - 29}{2 \cdot 1} : \frac{5 - 29}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 29}{2}, x = \frac{5 - 29}{2}

\frac{3}{5} m^{2} - 2 mn + \frac{1}{10} m^{2} - \frac{1}{3} mn + 2 mn - 2 m^{2}

3 (1 1^{\frac{1}{4}}) + 9 (1 1^{\frac{1}{4}})

3 (2)

\frac{7 x - 1}{3} \geq \frac{x + 2}{4}

25 \cdot 2

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(29))/2 ,x=(5-sqrt(29))/2
Find the zeros of x^2-5x-1=0
The zeros of x^2-5x-1=0 are x=(5+sqrt(29))/2 ,x=(5-sqrt(29))/2