(1-3x)(7x+1)-(1-3x)(2x-4)=0

Solution for (1-3x)(7x+1)-(1-3x)(2x-4)=0 equation:

(1-3x)(7x+1)-(1-3x)(2x-4)=0

We add all the numbers together, and all the variables

(-3x+1)(7x+1)-(-3x+1)(2x-4)=0

We multiply parentheses ..

(-21x^2-3x+7x+1)-(-3x+1)(2x-4)=0

We get rid of parentheses

-21x^2-3x+7x-(-3x+1)(2x-4)+1=0

We multiply parentheses ..

-21x^2-(-6x^2+12x+2x-4)-3x+7x+1=0

We add all the numbers together, and all the variables

-21x^2-(-6x^2+12x+2x-4)+4x+1=0

We get rid of parentheses

-21x^2+6x^2-12x-2x+4x+4+1=0

We add all the numbers together, and all the variables

-15x^2-10x+5=0

a = -15; b = -10; c = +5;
Δ = b2-4ac
Δ = -102-4·(-15)·5
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*-15}=\frac{-10}{-30}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*-15}=\frac{30}{-30}
=-1
$