What are the solutions to the equation 3(2-2x)=(x-4)(x-2) ?

The solutions to the equation 3(2-2x)=(x-4)(x-2) are x=sqrt(2)i,x=-sqrt(2)i

Find the zeros of 3(2-2x)=(x-4)(x-2)

The zeros of 3(2-2x)=(x-4)(x-2) are x=sqrt(2)i,x=-sqrt(2)i

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3 (2 - 2 x) = (x - 4) (x - 2)

x = 2 i, x = - 2 i

Solution steps

3 (2 - 2 x) = (x - 4) (x - 2)

Expand

3 (2 - 2 x) : 6 - 6 x

Expand

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

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

Switch sides

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

Move

8

to the right side

x^{2} - 6 x = - 6 x - 2

Move

6 x

to the left side

x^{2} = - 2

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - 2, x = - - 2

- 2 = 2 i

- - 2 = - 2 i

x = 2 i, x = - 2 i

x^{2} + 14 x + 40 = 0

101^{2} + x^{2} = 1 4^{2}

4 x^{2} + 77 = - 36

w, w^{2} + 4 w = 0

9 x - 2 - 2 x (3 x - 1) = 4 x^{2}

What are the solutions to the equation 3(2-2x)=(x-4)(x-2) ?
The solutions to the equation 3(2-2x)=(x-4)(x-2) are x=sqrt(2)i,x=-sqrt(2)i
Find the zeros of 3(2-2x)=(x-4)(x-2)
The zeros of 3(2-2x)=(x-4)(x-2) are x=sqrt(2)i,x=-sqrt(2)i