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

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

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

We move all terms to the left:

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

We multiply parentheses

2x^2+8x-(2*(x+2))+8=0


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

2*(x+2)

We multiply parentheses

2x+4

Back to the equation:

-(2x+4)


We get rid of parentheses

2x^2+8x-2x-4+8=0

We add all the numbers together, and all the variables

2x^2+6x+4=0

a = 2; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·2·4
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{-(6)-2}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*2}=\frac{-4}{4}
=-1
$