(2+X)=4+4x+x^2

Solution for (2+X)=4+4x+x^2 equation:

(2+X)=4+4X+X^2

We move all terms to the left:

(2+X)-(4+4X+X^2)=0

We add all the numbers together, and all the variables

-(4+4X+X^2)+(X+2)=0

We get rid of parentheses

-X^2-4X+X-4+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
$