(2x-1)(x+1)-(2x-1)(3x-5)=0

Solution for (2x-1)(x+1)-(2x-1)(3x-5)=0 equation:

(2x-1)(x+1)-(2x-1)(3x-5)=0

We multiply parentheses ..

(+2x^2+2x-1x-1)-(2x-1)(3x-5)=0

We get rid of parentheses

2x^2+2x-1x-(2x-1)(3x-5)-1=0

We multiply parentheses ..

2x^2-(+6x^2-10x-3x+5)+2x-1x-1=0

We add all the numbers together, and all the variables

2x^2-(+6x^2-10x-3x+5)+x-1=0

We get rid of parentheses

2x^2-6x^2+10x+3x+x-5-1=0

We add all the numbers together, and all the variables

-4x^2+14x-6=0

a = -4; b = 14; c = -6;
Δ = b2-4ac
Δ = 142-4·(-4)·(-6)
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{-(14)-10}{2*-4}=\frac{-24}{-8}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*-4}=\frac{-4}{-8}
=1/2
$