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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

x^2-(+2x^2-8x-3x+12)+2x-1x-1x-2+4=0

We add all the numbers together, and all the variables

x^2-(+2x^2-8x-3x+12)+2=0

We get rid of parentheses

x^2-2x^2+8x+3x-12+2=0

We add all the numbers together, and all the variables

-1x^2+11x-10=0

a = -1; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·(-1)·(-10)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*-1}=\frac{-2}{-2}
=1
$