(x+2)(x-5)=(x+1)(x+1)-10

Solution for (x+2)(x-5)=(x+1)(x+1)-10 equation:

(x+2)(x-5)=(x+1)(x+1)-10

We move all terms to the left:

(x+2)(x-5)-((x+1)(x+1)-10)=0

We multiply parentheses ..

(+x^2-5x+2x-10)-((x+1)(x+1)-10)=0


We calculate terms in parentheses: -((x+1)(x+1)-10), so:

(x+1)(x+1)-10

We multiply parentheses ..

(+x^2+x+x+1)-10

We get rid of parentheses

x^2+x+x+1-10

We add all the numbers together, and all the variables

x^2+2x-9

Back to the equation:

-(x^2+2x-9)


We get rid of parentheses

x^2-x^2-5x+2x-2x-10+9=0

We add all the numbers together, and all the variables

-5x-1=0

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

-5x=1

x=1/-5

x=-1/5