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

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

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

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

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

x = - 2, x = 8

Solution steps

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

Move

(x + 4) (x + 2)

to the left side

(2 x + 4) (x - 2) - (x + 4) (x + 2) = 0

Factor

(2 x + 4) (x - 2) - (x + 4) (x + 2) : (x + 2) (x - 8)

(x + 2) (x - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 8 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 8 = 0 : x = 8

The solutions to the quadratic equation are:

x = - 2, x = 8

(2 x - 1)^{2} - 225 = 0

3 x^{2} - 4 x = 3

x^{2} + 4 = 6 x - 5

49 x^{2} - 21 x + 2 = 0

2 x^{2} = \frac{a ^{3}}{2}

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