What are the solutions to the equation (x+2)^2=4-x ?

The solutions to the equation (x+2)^2=4-x are x=0,x=-5

Find the zeros of (x+2)^2=4-x

The zeros of (x+2)^2=4-x are x=0,x=-5

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

x = 0, x = - 5

Solution steps

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

Expand

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

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

Move

x

to the left side

x^{2} + 5 x + 4 = 4

Move

4

to the left side

x^{2} + 5 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 0 = 5

x_{1, 2} = \frac{- 5 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 5 + 5}{2 \cdot 1}, x_{2} = \frac{- 5 - 5}{2 \cdot 1}

x = \frac{- 5 + 5}{2 \cdot 1} : 0

x = \frac{- 5 - 5}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 0, x = - 5

(- 6 + 1) (x^{2} + 2 x) = (- 6 + 3) (3 x + 5)

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

30.8 \cdot 1 0^{5} = 0.9 (35) (7 5^{2}) (252) x (1 - 0.5 x)

25000 x^{2} + 600000 = 0

2 x^{2} + 8 x - 64 = 0

What are the solutions to the equation (x+2)^2=4-x ?
The solutions to the equation (x+2)^2=4-x are x=0,x=-5
Find the zeros of (x+2)^2=4-x
The zeros of (x+2)^2=4-x are x=0,x=-5