2(x+2)(x+4)=28

Solution for 2(x+2)(x+4)=28 equation:

2(x+2)(x+4)=28

We move all terms to the left:

2(x+2)(x+4)-(28)=0

We multiply parentheses ..

2(+x^2+4x+2x+8)-28=0

We multiply parentheses

2x^2+8x+4x+16-28=0

We add all the numbers together, and all the variables

2x^2+12x-12=0

a = 2; b = 12; c = -12;
Δ = b2-4ac
Δ = 122-4·2·(-12)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{15}}{2*2}=\frac{-12-4\sqrt{15}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{15}}{2*2}=\frac{-12+4\sqrt{15}}{4}
$