What are the solutions to the equation x^2-4x-1=0 ?

The solutions to the equation x^2-4x-1=0 are x=2+sqrt(5),x=2-sqrt(5)

Find the zeros of x^2-4x-1=0

The zeros of x^2-4x-1=0 are x=2+sqrt(5),x=2-sqrt(5)

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x^{2} - 4 x - 1 = 0

x = 2 + 5, x = 2 - 5

Solution steps

x^{2} - 4 x - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 1) = 25

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 5}{2 \cdot 1} : 2 + 5

x = \frac{- ( - 4 ) - 2 5}{2 \cdot 1} : 2 - 5

The solutions to the quadratic equation are:

x = 2 + 5, x = 2 - 5

0.6 x^{2} - 2.4 x - 0.6 = 0

\frac{x ^{- 6}}{x ^{- 4}}

0 = x^{2} - 41 x + 238

\frac{- 9 x y ^{5} z ^{3}}{36 x y ^{4} z ^{5}}

5 x (8 - x) = 10

What are the solutions to the equation x^2-4x-1=0 ?
The solutions to the equation x^2-4x-1=0 are x=2+sqrt(5),x=2-sqrt(5)
Find the zeros of x^2-4x-1=0
The zeros of x^2-4x-1=0 are x=2+sqrt(5),x=2-sqrt(5)