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

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

(X+1)(X-3)+(X+1)(X+1)=2X(X-4)

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(2X(X-4)), so:

2X(X-4)

We multiply parentheses

2X^2-8X

Back to the equation:

-(2X^2-8X)


We get rid of parentheses

X^2-2X^2-3X+X+(X+1)(X+1)+8X-3=0

We multiply parentheses ..

X^2-2X^2+(+X^2+X+X+1)-3X+X+8X-3=0

We add all the numbers together, and all the variables

-1X^2+(+X^2+X+X+1)+6X-3=0

We get rid of parentheses

-1X^2+X^2+X+X+6X+1-3=0

We add all the numbers together, and all the variables

8X-2=0

We move all terms containing X to the left, all other terms to the right

8X=2

X=2/8

X=1/4