What are the solutions to the equation (2x+2)^2+5=1 ?

The solutions to the equation (2x+2)^2+5=1 are x=i-1,x=-1-i

Find the zeros of (2x+2)^2+5=1

The zeros of (2x+2)^2+5=1 are x=i-1,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 + 2)^{2} + 5 = 1

x = i - 1, x = - 1 - i

Solution steps

(2 x + 2)^{2} + 5 = 1

Move

5

to the right side

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

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

2 x + 2 = - 4 : x = i - 1

Solve

2 x + 2 = - - 4 : x = - 1 - i

The solutions to the quadratic equation are:

x = i - 1, x = - 1 - i

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

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

3 t^{2} - 24 t + 21 = 0

x^{2} + 26 x + 167.5 = 0

I, P = I^{2} R

What are the solutions to the equation (2x+2)^2+5=1 ?
The solutions to the equation (2x+2)^2+5=1 are x=i-1,x=-1-i
Find the zeros of (2x+2)^2+5=1
The zeros of (2x+2)^2+5=1 are x=i-1,x=-1-i