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

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

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

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

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

x = 1 - 2, x = 1 + 2

Decimal Notation

x = - 0.41421 \dots, x = 2.41421 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 1 = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - 2, x = 1 + 2

x + 16 \leq 25

\frac{c ^{4}}{c ^{8}}

∣ 2 v + 3 ∣ = ∣ 2 v + 6 ∣

16!

\frac{1.32 \cdot 1 0 ^{5}}{2 \cdot 1 0 ^{- 3}}

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