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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-1)(x+5)-(3x+2)(x-1)), so:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

x^2-(+3x^2-3x+2x-2)+4x-5

We get rid of parentheses

x^2-3x^2+3x-2x+4x+2-5

We add all the numbers together, and all the variables

-2x^2+5x-3

Back to the equation:

-(-2x^2+5x-3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2-10x=0

a = 4; b = -10; c = 0;
Δ = b2-4ac
Δ = -102-4·4·0
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{-(-10)-10}{2*4}=\frac{0}{8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2*4}=\frac{20}{8}
=2+1/2
$