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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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


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

(+x^2+2x+x+2)

We get rid of parentheses

x^2+2x+x+2

We add all the numbers together, and all the variables

x^2+3x+2

Back to the equation:

-(x^2+3x+2)


We add all the numbers together, and all the variables

2x-(x^2+3x+2)-5=0

We get rid of parentheses

-x^2+2x-3x-2-5=0

We add all the numbers together, and all the variables

-1x^2-1x-7=0

a = -1; b = -1; c = -7;
Δ = b2-4ac
Δ = -12-4·(-1)·(-7)
Δ = -27
Delta is less than zero, so there is no solution for the equation