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=sqrt(5)-2

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

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

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

x = - 2 - 5, x = 5 - 2

Decimal Notation

x = - 4.23606 \dots, x = 0.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2 - 5, x = 5 - 2

∣ x - 4 ∣ + 3 = 9

7 x^{2} - 112 = 0

(47 + 3) (67 - 1)

x^{3} + 5 x^{2} + 8 x + 4

7 (3 x + 9) = 11 - (x + 3)

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