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 ( - 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 )} : 2 + 5

The solutions to the quadratic equation are:

x = 2 - 5, x = 2 + 5

64 x^{2} - 64 x + 7

(5 - 2 i) (2 + 7 i) - 2 i (3 i)^{2}

0.72 x = 252

20 x^{8} y^{2} z^{2} + 15 x^{5} y^{2} z + 35 x^{3} y^{3} z

2 (x - 2) = 2 (x - 2)

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)