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

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

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

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

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

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

Solution steps

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

Expand

(x + 2) (x + 4) : 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

1 + (- \frac{2}{3}) - (- m) m = \frac{9}{2}

x^{2} = - 56

R, 200 = \frac{1}{3} π R^{2} H

2 x^{2} + 7 x - 184 = 0

8 x^{2} - 23 x - 3 = 0

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