(2x+3)(x-1)-(x+4)(x-2)=9-x

Solution for (2x+3)(x-1)-(x+4)(x-2)=9-x equation:

(2x+3)(x-1)-(x+4)(x-2)=9-x

We move all terms to the left:

(2x+3)(x-1)-(x+4)(x-2)-(9-x)=0

We add all the numbers together, and all the variables

(2x+3)(x-1)-(x+4)(x-2)-(-1x+9)=0

We get rid of parentheses

(2x+3)(x-1)-(x+4)(x-2)+1x-9=0

We multiply parentheses ..

(+2x^2-2x+3x-3)-(x+4)(x-2)+1x-9=0

We add all the numbers together, and all the variables

(+2x^2-2x+3x-3)+x-(x+4)(x-2)-9=0

We get rid of parentheses

2x^2-2x+3x+x-(x+4)(x-2)-3-9=0

We multiply parentheses ..

2x^2-(+x^2-2x+4x-8)-2x+3x+x-3-9=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-x^2+2x-4x+2x+8-12=0

We add all the numbers together, and all the variables

x^2-4=0

a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2}
=2
$