165=(8+2x)(4+2x)-32

Solution for 165=(8+2x)(4+2x)-32 equation:

165=(8+2x)(4+2x)-32

We move all terms to the left:

165-((8+2x)(4+2x)-32)=0

We add all the numbers together, and all the variables

-((2x+8)(2x+4)-32)+165=0

We multiply parentheses ..

-((+4x^2+8x+16x+32)-32)+165=0


We calculate terms in parentheses: -((+4x^2+8x+16x+32)-32), so:

(+4x^2+8x+16x+32)-32

We get rid of parentheses

4x^2+8x+16x+32-32

We add all the numbers together, and all the variables

4x^2+24x

Back to the equation:

-(4x^2+24x)


We get rid of parentheses

-4x^2-24x+165=0

a = -4; b = -24; c = +165;
Δ = b2-4ac
Δ = -242-4·(-4)·165
Δ = 3216
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:

$\sqrt{\Delta}=\sqrt{3216}=\sqrt{16*201}=\sqrt{16}*\sqrt{201}=4\sqrt{201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{201}}{2*-4}=\frac{24-4\sqrt{201}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{201}}{2*-4}=\frac{24+4\sqrt{201}}{-8}
$