(8+2x)(4+2x)-32=32

Solution for (8+2x)(4+2x)-32=32 equation:

(8+2x)(4+2x)-32=32

We move all terms to the left:

(8+2x)(4+2x)-32-(32)=0

We add all the numbers together, and all the variables

(2x+8)(2x+4)-32-32=0

We add all the numbers together, and all the variables

(2x+8)(2x+4)-64=0

We multiply parentheses ..

(+4x^2+8x+16x+32)-64=0

We get rid of parentheses

4x^2+8x+16x+32-64=0

We add all the numbers together, and all the variables

4x^2+24x-32=0

a = 4; b = 24; c = -32;
Δ = b2-4ac
Δ = 242-4·4·(-32)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{17}}{2*4}=\frac{-24-8\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{17}}{2*4}=\frac{-24+8\sqrt{17}}{8}
$