(16+2x)(12+2x)=96

Solution for (16+2x)(12+2x)=96 equation:

(16+2x)(12+2x)=96

We move all terms to the left:

(16+2x)(12+2x)-(96)=0

We add all the numbers together, and all the variables

(2x+16)(2x+12)-96=0

We multiply parentheses ..

(+4x^2+24x+32x+192)-96=0

We get rid of parentheses

4x^2+24x+32x+192-96=0

We add all the numbers together, and all the variables

4x^2+56x+96=0

a = 4; b = 56; c = +96;
Δ = b2-4ac
Δ = 562-4·4·96
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-40}{2*4}=\frac{-96}{8}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+40}{2*4}=\frac{-16}{8}
=-2
$