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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-2)·6
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-2}=\frac{-12}{-4}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-2}=\frac{4}{-4}
=-1
$