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

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

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

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

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

x = 1

Solution steps

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

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

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

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

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

x = 1

The solution to the quadratic equation is:

x = 1

1.2 = \frac{x}{3} (x - 6) + 4

(2 x + 8) (5 x + 2) = 0

y, x^{3} + 5 x^{2} y + 2 y^{2} = 4 y + 11

6 X^{2} + 11 X - 10 = 0

(2 x + 5)^{2} = 25

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