What are the solutions to the equation (x+2)^2=8+x ?

The solutions to the equation (x+2)^2=8+x are x=1,x=-4

Find the zeros of (x+2)^2=8+x

The zeros of (x+2)^2=8+x are x=1,x=-4

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(x + 2)^{2} = 8 + x

x = 1, x = - 4

Solution steps

(x + 2)^{2} = 8 + x

Expand

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

x^{2} + 4 x + 4 = 8 + x

Move

x

to the left side

x^{2} + 3 x + 4 = 8

Move

8

to the left side

x^{2} + 3 x - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

x = \frac{- 3 + 5}{2 \cdot 1} : 1

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

x^{2} - 5 x + 10

(2 x - 5)^{2} = 7

4 x^{2} + 1 = 7

2 x^{2} + 15 x - 23 = 2 x

(r - 2) (r + 8) = - 5

What are the solutions to the equation (x+2)^2=8+x ?
The solutions to the equation (x+2)^2=8+x are x=1,x=-4
Find the zeros of (x+2)^2=8+x
The zeros of (x+2)^2=8+x are x=1,x=-4