4x+x(x-8)=(x+1)(x-2)-7

Solution for 4x+x(x-8)=(x+1)(x-2)-7 equation:

4x+x(x-8)=(x+1)(x-2)-7

We move all terms to the left:

4x+x(x-8)-((x+1)(x-2)-7)=0

We multiply parentheses

x^2+4x-8x-((x+1)(x-2)-7)=0

We multiply parentheses ..

x^2-((+x^2-2x+x-2)-7)+4x-8x=0


We calculate terms in parentheses: -((+x^2-2x+x-2)-7), so:

(+x^2-2x+x-2)-7

We get rid of parentheses

x^2-2x+x-2-7

We add all the numbers together, and all the variables

x^2-1x-9

Back to the equation:

-(x^2-1x-9)


We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-x^2-4x+1x+9=0

We add all the numbers together, and all the variables

-3x+9=0

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

-3x=-9

x=-9/-3

x=+3