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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2+x+x+1)), so:

(+x^2+x+x+1)

We get rid of parentheses

x^2+x+x+1

We add all the numbers together, and all the variables

x^2+2x+1

Back to the equation:

-(x^2+2x+1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·(-1)·1
Δ = 5
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{5}}{2*-1}=\frac{1-\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{5}}{2*-1}=\frac{1+\sqrt{5}}{-2}
$