(12+2x)(4+2x)=96

Solution for (12+2x)(4+2x)=96 equation:

(12+2x)(4+2x)=96

We move all terms to the left:

(12+2x)(4+2x)-(96)=0

We add all the numbers together, and all the variables

(2x+12)(2x+4)-96=0

We multiply parentheses ..

(+4x^2+8x+24x+48)-96=0

We get rid of parentheses

4x^2+8x+24x+48-96=0

We add all the numbers together, and all the variables

4x^2+32x-48=0

a = 4; b = 32; c = -48;
Δ = b2-4ac
Δ = 322-4·4·(-48)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{7}}{2*4}=\frac{-32-16\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{7}}{2*4}=\frac{-32+16\sqrt{7}}{8}
$