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

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

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

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

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

x = 1, x = - 4

Solution steps

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

Expand

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

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

Move

8

to the left side

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

Move

x

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} - 2 x - 3 = 5

3 x^{2} - 12 x + 6 = 0

x^{2} - 4 x = - 12

x^{2} + 3.9 x + 1.8 = 0

a x^{2} + b x + 1 = 0

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