What are the solutions to the equation x^2+2x+6=2(x+1) ?

The solutions to the equation x^2+2x+6=2(x+1) are x=2i,x=-2i

Find the zeros of x^2+2x+6=2(x+1)

The zeros of x^2+2x+6=2(x+1) are x=2i,x=-2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 2 x + 6 = 2 (x + 1)

x = 2 i, x = - 2 i

Solution steps

x^{2} + 2 x + 6 = 2 (x + 1)

Expand

2 (x + 1) : 2 x + 2

x^{2} + 2 x + 6 = 2 x + 2

Move

6

to the right side

x^{2} + 2 x = 2 x - 4

Move

2 x

to the left side

x^{2} = - 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - 4, x = - - 4

- 4 = 2 i

- - 4 = - 2 i

x = 2 i, x = - 2 i

(x) = 1500 + 5 x + 0.01 x^{2}

3 x^{2} + 18 x - 6

(x - 1)^{2} = (4 x - 7)^{2} + (3 x)^{2}

2 (x + 3)^{2} = 90

t + t^{2} = 0

What are the solutions to the equation x^2+2x+6=2(x+1) ?
The solutions to the equation x^2+2x+6=2(x+1) are x=2i,x=-2i
Find the zeros of x^2+2x+6=2(x+1)
The zeros of x^2+2x+6=2(x+1) are x=2i,x=-2i