2(x^2+6x+8)=4(x^2+4x+4)

Solution for 2(x^2+6x+8)=4(x^2+4x+4) equation:

2(x^2+6x+8)=4(x^2+4x+4)

We move all terms to the left:

2(x^2+6x+8)-(4(x^2+4x+4))=0

We multiply parentheses

2x^2+12x-(4(x^2+4x+4))+16=0


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

4(x^2+4x+4)

We multiply parentheses

4x^2+16x+16

Back to the equation:

-(4x^2+16x+16)


We get rid of parentheses

2x^2-4x^2+12x-16x-16+16=0

We add all the numbers together, and all the variables

-2x^2-4x=0

a = -2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-2)·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*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-2}=\frac{8}{-4}
=-2
$