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

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

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

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

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

x = - 1, x = - 2

Solution steps

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

Expand

(x + 2)^{2} : x^{2} + 4 x + 4

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

Move

2

to the left side

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

Move

x

to the left side

x^{2} + 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot 2 = 1

x_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 1}{2 \cdot 1}, x_{2} = \frac{- 3 - 1}{2 \cdot 1}

x = \frac{- 3 + 1}{2 \cdot 1} : - 1

x = \frac{- 3 - 1}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = - 1, x = - 2

(\frac{9}{41})^{2} + y^{2} = 1

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

13 x^{2} - 3 x = 4 x^{2} + 21

z, g = k (z - 0.8) (z + 0.7)

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

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