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

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

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

We move all terms to the left:

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

We get rid of parentheses

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


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

(x^2+2x+1)

We get rid of parentheses

x^2+2x+1

Back to the equation:

-(x^2+2x+1)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-1x+4=0

a = -1; b = -1; c = +4;
Δ = b2-4ac
Δ = -12-4·(-1)·4
Δ = 17
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{17}}{2*-1}=\frac{1-\sqrt{17}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{17}}{2*-1}=\frac{1+\sqrt{17}}{-2}
$