(2x-2)+(x+1)+x+(x+1)=(2x-9)(x+1)+(x+8)+x

Solution for (2x-2)+(x+1)+x+(x+1)=(2x-9)(x+1)+(x+8)+x equation:

(2x-2)+(x+1)+x+(x+1)=(2x-9)(x+1)+(x+8)+x

We move all terms to the left:

(2x-2)+(x+1)+x+(x+1)-((2x-9)(x+1)+(x+8)+x)=0

We add all the numbers together, and all the variables

x+(2x-2)+(x+1)+(x+1)-((2x-9)(x+1)+(x+8)+x)=0

We get rid of parentheses

x+2x+x+x-((2x-9)(x+1)+(x+8)+x)-2+1+1=0

We multiply parentheses ..

-((+2x^2+2x-9x-9)+(x+8)+x)+x+2x+x+x-2+1+1=0


We calculate terms in parentheses: -((+2x^2+2x-9x-9)+(x+8)+x), so:

(+2x^2+2x-9x-9)+(x+8)+x

We add all the numbers together, and all the variables

(+2x^2+2x-9x-9)+x+(x+8)

We get rid of parentheses

2x^2+2x-9x+x+x-9+8

We add all the numbers together, and all the variables

2x^2-5x-1

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

-2x^2+5x+5x+1=0

We add all the numbers together, and all the variables

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

a = -2; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·(-2)·1
Δ = 108
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{3}}{2*-2}=\frac{-10-6\sqrt{3}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{3}}{2*-2}=\frac{-10+6\sqrt{3}}{-4}
$