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

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

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

We move all terms to the left:

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

We multiply parentheses

(x+3)(x+4)-6x-((x-4))+4=0

We multiply parentheses ..

(+x^2+4x+3x+12)-6x-((x-4))+4=0


We calculate terms in parentheses: -((x-4)), so:

(x-4)

We get rid of parentheses

x-4

Back to the equation:

-(x-4)


We get rid of parentheses

x^2+4x+3x-6x-x+12+4+4=0

We add all the numbers together, and all the variables

x^2+20=0

a = 1; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·1·20
Δ = -80
Delta is less than zero, so there is no solution for the equation