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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3x+12x-(2x(x+1))-4=0


We calculate terms in parentheses: -(2x(x+1)), so:

2x(x+1)

We multiply parentheses

2x^2+2x

Back to the equation:

-(2x^2+2x)


We add all the numbers together, and all the variables

15x-(2x^2+2x)-4=0

We get rid of parentheses

-2x^2+15x-2x-4=0

We add all the numbers together, and all the variables

-2x^2+13x-4=0

a = -2; b = 13; c = -4;
Δ = b2-4ac
Δ = 132-4·(-2)·(-4)
Δ = 137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{137}}{2*-2}=\frac{-13-\sqrt{137}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{137}}{2*-2}=\frac{-13+\sqrt{137}}{-4}
$