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

Solution for 1=(x^2+4x+4) equation:

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

We move all terms to the left:

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


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

(x^2+4x+4)

We get rid of parentheses

x^2+4x+4

Back to the equation:

-(x^2+4x+4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2-4x-3=0

a = -1; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·(-1)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*-1}=\frac{2}{-2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*-1}=\frac{6}{-2}
=-3
$