2209=x^2+(x^2+4x+4)

Solution for 2209=x^2+(x^2+4x+4) equation:

2209=x^2+(x^2+4x+4)

We move all terms to the left:

2209-(x^2+(x^2+4x+4))=0


We calculate terms in parentheses: -(x^2+(x^2+4x+4)), so:

x^2+(x^2+4x+4)

We get rid of parentheses

x^2+x^2+4x+4

We add all the numbers together, and all the variables

2x^2+4x+4

Back to the equation:

-(2x^2+4x+4)


We get rid of parentheses

-2x^2-4x-4+2209=0

We add all the numbers together, and all the variables

-2x^2-4x+2205=0

a = -2; b = -4; c = +2205;
Δ = b2-4ac
Δ = -42-4·(-2)·2205
Δ = 17656
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{17656}=\sqrt{4*4414}=\sqrt{4}*\sqrt{4414}=2\sqrt{4414}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{4414}}{2*-2}=\frac{4-2\sqrt{4414}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{4414}}{2*-2}=\frac{4+2\sqrt{4414}}{-4}
$