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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

x^2+2x+2x+4

We add all the numbers together, and all the variables

x^2+4x+4

Back to the equation:

-(x^2+4x+4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-4x=0

a = -1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-1)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-1}=\frac{0}{-2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-1}=\frac{8}{-2}
=-4
$