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

The solutions to the equation (x+4)(x+2)=2x are x=-2+2i,x=-2-2i

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

The zeros of (x+4)(x+2)=2x are x=-2+2i,x=-2-2i

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(x + 4) (x + 2) = 2 x

x = - 2 + 2 i, x = - 2 - 2 i

Solution steps

(x + 4) (x + 2) = 2 x

Expand

(x + 4) (x + 2) : x^{2} + 6 x + 8

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

Move

2 x

to the left side

x^{2} + 4 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 8 : 4 i

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

Separate the solutions

x_{1} = \frac{- 4 + 4 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 i}{2 \cdot 1}

x = \frac{- 4 + 4 i}{2 \cdot 1} : - 2 + 2 i

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

The solutions to the quadratic equation are:

x = - 2 + 2 i, x = - 2 - 2 i

p (3 p + 20) = 7

2 x^{2} - 12 x + 21

5 x^{2} + x - 16 = - 10

- 0.25 x^{2} + 4 x - 12 = 0

k^{2} + k - 42 = 0

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