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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(3(-1x+1)-4), so:

3(-1x+1)-4

We multiply parentheses

-3x+3-4

We add all the numbers together, and all the variables

-3x-1

Back to the equation:

-(-3x-1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-4x^2-12x+3x+1+3=0

We add all the numbers together, and all the variables

-4x^2-9x+4=0

a = -4; b = -9; c = +4;
Δ = b2-4ac
Δ = -92-4·(-4)·4
Δ = 145
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{-(-9)-\sqrt{145}}{2*-4}=\frac{9-\sqrt{145}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{145}}{2*-4}=\frac{9+\sqrt{145}}{-8}
$