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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -6; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-6)·4
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*-6}=\frac{-12}{-12}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*-6}=\frac{8}{-12}
=-2/3
$