3*(2x-4)=7*(x+2)x+2

Solution for 3*(2x-4)=7*(x+2)x+2 equation:

3(2x-4)=7(x+2)x+2

We move all terms to the left:

3(2x-4)-(7(x+2)x+2)=0

We multiply parentheses

6x-(7(x+2)x+2)-12=0


We calculate terms in parentheses: -(7(x+2)x+2), so:

7(x+2)x+2

We multiply parentheses

7x^2+14x+2

Back to the equation:

-(7x^2+14x+2)


We get rid of parentheses

-7x^2+6x-14x-2-12=0

We add all the numbers together, and all the variables

-7x^2-8x-14=0

a = -7; b = -8; c = -14;
Δ = b2-4ac
Δ = -82-4·(-7)·(-14)
Δ = -328
Delta is less than zero, so there is no solution for the equation