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

Solution for 0=(x+2)(x+1) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

x^2+x+2x+2

We add all the numbers together, and all the variables

x^2+3x+2

Back to the equation:

-(x^2+3x+2)


We get rid of parentheses

-x^2-3x-2=0

We add all the numbers together, and all the variables

-1x^2-3x-2=0

a = -1; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·(-1)·(-2)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*-1}=\frac{2}{-2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-1}=\frac{4}{-2}
=-2
$