What are the solutions to the equation (x+3)^2-2=2x(x+2)+x+8 ?

The solutions to the equation (x+3)^2-2=2x(x+2)+x+8 are x= 1/2-i(sqrt(3))/2 ,x= 1/2+i(sqrt(3))/2

Find the zeros of (x+3)^2-2=2x(x+2)+x+8

The zeros of (x+3)^2-2=2x(x+2)+x+8 are x= 1/2-i(sqrt(3))/2 ,x= 1/2+i(sqrt(3))/2

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(x + 3)^{2} - 2 = 2 x (x + 2) + x + 8

x = \frac{1}{2} - i \frac{3}{2}, x = \frac{1}{2} + i \frac{3}{2}

Solution steps

(x + 3)^{2} - 2 = 2 x (x + 2) + x + 8

Expand

(x + 3)^{2} - 2 : x^{2} + 6 x + 7

Expand

2 x (x + 2) + x + 8 : 2 x^{2} + 5 x + 8

x^{2} + 6 x + 7 = 2 x^{2} + 5 x + 8

Move

8

to the left side

x^{2} + 6 x - 1 = 2 x^{2} + 5 x

Move

5 x

to the left side

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

Move

2 x^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

1^{2} - 4 (- 1) (- 1) : 3 i

x_{1, 2} = \frac{- 1 \pm 3 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 3 i}{2 ( - 1 )}, x_{2} = \frac{- 1 - 3 i}{2 ( - 1 )}

x = \frac{- 1 + 3 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{3}{2}

x = \frac{- 1 - 3 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} - i \frac{3}{2}, x = \frac{1}{2} + i \frac{3}{2}

x^{2} + 71 = 18 x - 9

m^{2} - 3 m + 5 = 0

(y + 7)^{2} = 2 y^{2} + 17 + 39

x, y^{2} - x^{2} = c

3 x^{2} - 7 x + 2 = 0

What are the solutions to the equation (x+3)^2-2=2x(x+2)+x+8 ?
The solutions to the equation (x+3)^2-2=2x(x+2)+x+8 are x= 1/2-i(sqrt(3))/2 ,x= 1/2+i(sqrt(3))/2
Find the zeros of (x+3)^2-2=2x(x+2)+x+8
The zeros of (x+3)^2-2=2x(x+2)+x+8 are x= 1/2-i(sqrt(3))/2 ,x= 1/2+i(sqrt(3))/2