44=(2x+12)(2x+8)

Solution for 44=(2x+12)(2x+8) equation:

44=(2x+12)(2x+8)

We move all terms to the left:

44-((2x+12)(2x+8))=0

We multiply parentheses ..

-((+4x^2+16x+24x+96))+44=0


We calculate terms in parentheses: -((+4x^2+16x+24x+96)), so:

(+4x^2+16x+24x+96)

We get rid of parentheses

4x^2+16x+24x+96

We add all the numbers together, and all the variables

4x^2+40x+96

Back to the equation:

-(4x^2+40x+96)


We get rid of parentheses

-4x^2-40x-96+44=0

We add all the numbers together, and all the variables

-4x^2-40x-52=0

a = -4; b = -40; c = -52;
Δ = b2-4ac
Δ = -402-4·(-4)·(-52)
Δ = 768
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:

$\sqrt{\Delta}=\sqrt{768}=\sqrt{256*3}=\sqrt{256}*\sqrt{3}=16\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{3}}{2*-4}=\frac{40-16\sqrt{3}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{3}}{2*-4}=\frac{40+16\sqrt{3}}{-8}
$