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

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



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-7x-5=0

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

-7x=5

x=5/-7

x=-5/7

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