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

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

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

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

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

x = 0, x = 1

Solution steps

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

Move

1

to the left side

- x^{2} + x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

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

Separate the solutions

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

x = \frac{- 1 + 1}{2 ( - 1 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 1

- 4 x + \frac{5}{3} = \frac{7}{3}

\frac{- 3}{- 3 + 1} - 1

8 - 2 x^{2} < 4

8 x^{3} - 12 x

6 x - 15 = 6 x + 15

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