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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x - x^{2} = 2

x = 1 - i, x = 1 + i

Solution steps

2 x - x^{2} = 2

Move

2

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 (- 1) (- 2) : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 - i, x = 1 + i

0.25 t = 0.25 - t

x^{2} = 48

- 6 + 4 i

∣ x - 1 ∣ \geq 6

- 1.2 (1)^{2} + 9.6 (1)

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