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(3),x=-2+sqrt(3)

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

The zeros of -x^2-4x-1=0 are x=-2-sqrt(3),x=-2+sqrt(3)

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

x = - 2 - 3, x = - 2 + 3

Decimal Notation

x = - 3.73205 \dots, x = - 0.26794 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) (- 1) = 23

x_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3}{2 ( - 1 )}, x_{2} = \frac{- ( - 4 ) - 2 3}{2 ( - 1 )}

x = \frac{- ( - 4 ) + 2 3}{2 ( - 1 )} : - 2 - 3

x = \frac{- ( - 4 ) - 2 3}{2 ( - 1 )} : - 2 + 3

The solutions to the quadratic equation are:

x = - 2 - 3, x = - 2 + 3

1 0^{4}

(\frac{1}{2})^{4}

5 - 2 x \geq - \frac{1}{4}

5 n^{2} - 15

28 + 19 u = - 8 + 15 u

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(3),x=-2+sqrt(3)
Find the zeros of -x^2-4x-1=0
The zeros of -x^2-4x-1=0 are x=-2-sqrt(3),x=-2+sqrt(3)