3(x-4)+1=2(x+3)x=

Solution for 3(x-4)+1=2(x+3)x= equation:

3(x-4)+1=2(x+3)x=

We move all terms to the left:

3(x-4)+1-(2(x+3)x)=0

We multiply parentheses

3x-(2(x+3)x)-12+1=0


We calculate terms in parentheses: -(2(x+3)x), so:

2(x+3)x

We multiply parentheses

2x^2+6x

Back to the equation:

-(2x^2+6x)


We add all the numbers together, and all the variables

3x-(2x^2+6x)-11=0

We get rid of parentheses

-2x^2+3x-6x-11=0

We add all the numbers together, and all the variables

-2x^2-3x-11=0

a = -2; b = -3; c = -11;
Δ = b2-4ac
Δ = -32-4·(-2)·(-11)
Δ = -79
Delta is less than zero, so there is no solution for the equation